# NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions

NCERT solutions for class 12 chemistry chapter 2 Solutions- In our daily life we come across various mixtures like soft drinks, syrups and air. All of them are mixtures of two or more pure substances like air is a mixture of mainly nitrogen and oxygen etc. Also, you know about various types of mixtures or solutions like gaseous solutions, liquid solutions and solid solutions. NCERT solutions for class 12 chemistry chapter 2 Solutions mainly discuss questions based on liquid solutions and their properties. The solutions of NCERT class 12 chemistry chapter 2 Solutions also cover other questions based on important concepts like types of solutions, Raoult's law and Henry's law, the concentration of solutions in different units, solubility, vapour pressure of liquid solutions, ideal and non-ideal solutions, colligative properties, determination of molar mass and abnormal molar masses. These topics of this chapter are not only important for the class 12 CBSE Board exam but also for the various competitive exams like JEE Mains, NEET, BITSAT, VITEEE, etc. In CBSE NCERT solutions for class 12 chemistry chapter 2 Solutions, there are direct answers to the 41 questions which are there in the chapter exercise. To develop a grip on the topic, NCERT solutions for class 12 chemistry chapter 2 Solutions are prepared by chemistry experts in a very comprehensive manner. The students will be able to find step-by-step solutions which will eventually help you to write good answers and get good marks in the CBSE exam. The NCERT solutions which are provided here are free of cost and are easily accessible and if you wish to check other classes solutions, you can just click on the above link.

What are the solutions? - The solutions are mixtures of two or more than two components and solutions are classified as solids, liquids, and gaseous solution. The concentration of the solution is expressed in terms of molarity, molality, mole fraction and in percentages.

 $Molality(m)=\frac{Number \:of \:moles \:of \:solute}{Weight\:of\:solvent(kg)}$ $Molarity(N)=\frac{Number \:of \:moles \:of \:solute}{Volume\:of\:solution(kg)}$ $Normality(N)=\frac{Number \:of \:gram\:equivalent \:of \:solute}{Volume\:of\:solution(kg)}$

Topics and Sub-topics of NCERT Class 12 Chemistry Chapter 2 Solutions-

2.1Types of Solutions

2.2 Expressing Concentration of Solutions

2.3 Solubility

2.4 Vapour Pressure of Liquid Solutions

2.5 Ideal and Non-ideal Solutions

2.6 Colligative Properties and Determination of Molar Mass

2.7 Abnormal Molar Masses

## Solutions to In-Text Questions Exercise 2.1  to 2.12

### Question 2.1   Calculate the mass percentage of benzene $\inline (C_{6}H_{6})$  and carbon tetrachloride $\inline (CCl_{4})$  if $\inline 22\; g$ of benzene is dissolved in $\inline 122\; g$ of carbon tetrachloride.

We know that solute and solvent forms solution.

So mass percentage of benzene (solute)   :-

$=\frac{22}{22+122}\times100 = \frac{22}{144}\times100 = 15.28\%$

Similarly mass percentage of CCl4 :-

$=\frac{122}{22+122}\times100 = \frac{122}{144}\times100 = 84.72\%$

### Question 2.2   Calculate the mole fraction of benzene in solution containing $30\%$ by mass in carbon tetrachloride.

For calculating mole fraction, we need moles of both the compounds.

It is given that benzene is $30\%$ in the solution by mass.

So if we consider 100g of solution then 30g is benzene and 70g is CCl4.

$Moles of CCl_{4}= \frac{Given\ mass}{Molar\ mass} = \frac{70}{154} = 0.4545\ mol$

$\left ( Molar\ mass\ of\ CCl_4 = 12 + 4(35.5) \right )$

Similarly moles of benzene :

$= \frac{30}{78} = 0.3846$                                                                                    $\left ( Molar\ mass\ of\ benzene = 6(12) + 6(1) \right )$

So mole fraction of benzene is given  :

$= \frac{0.3846}{0.3846+0.4545}$

$= 0.458$

## Question 2.3(a)   Calculate the molarity of each of the following solutions:

$\inline 30\; g$ of  $\inline Co(NO_{3})_{2}.\; 6H_{2}O$  in  $\inline 4.3\: L$  of solution

For finding molarity we need the moles of solute and volume of solution.

So moles of solute :

$= \frac{Given\ mass}{Molar\ mass} = \frac{30}{291} = 0.103$

$Since\ molar\ mass\ of\ Co(NO_3)_2.6H_2O \right = 59 + 2(14 + 3 � 16) + 6 (16 +1 � 2)$

$= 291\ g\ mol^{-1}$

Now,                                     $Molarity = \frac{No.\ of\ moles\ of\ solute}{Volume\ of\ solution\ (l)}$

$= \frac{0.103}{4.3} = 0.023\ M$

### Question 2.3(b)   Calculate the molarity of each of the following solutions:

$\inline 30\; mL$  of $\inline 0.5\; M$  $\inline H_{2}SO_{4}$  diluted to $\inline 500 \; mL$.

By conservation of moles we can write :

M1V= M2V2

Given that M1 = 0.5 M and V1 =  30 ml ;   V2 =  500 ml

$M_2 = \frac{M_1V_1}{V_2} = 0.03\ M$

## Question 2.4   Calculate the mass of urea  $\inline (NH_{2}CONH_{2})$  required in making  $\inline 2.5 \; kg$  of  $\inline 0.25\; molal$  aqueous solution.

Let us assume that the mass of urea required be x g.

So moles of urea will be :

$Moles = \frac{Given\ mass}{Molar\ mass} = \frac{x}{60}\ moles$

$Molality = \frac{Moles}{Mass\ of\ solvent\ in\ Kg} = \frac{\frac{x}{60}}{2.5-0.001x} = 0.25$

we get      x =  37

Thus mass of urea required = 37 g.

### Question 2.5   Calculate

(a) molality

(b) molarity and

(c) mole fraction

of KI if the density of $20\%$ (mass/mass) aqueous KI is $\inline 1.202\; g\; mL^{-1}$.

If we assume our solution is 100 g. Then according to question,  20 g KI is present and 80 g is water.

So moles of KI   :

$=\frac{20}{166}$           $\left ( Molar\ mass = 39+127 = 166\ g\ mol^{-1} \right )$

(a) Molality :-

$Molality = \frac{Moles}{Mass\ of\ solvent\ in\ Kg} = \frac{\frac{20}{166}}{0.08}= 1.506\ m.$

(b) Molarity :-

$Density = \frac{Mass}{Volume}$

$Volume = \frac{Mass}{Density} = \frac{100}{1.202} = 83.19 mL$

$Molarity = \frac{Moles}{Volume(l)} = \frac{\frac{20}{166}}{83.19\times10^{-3}} = 1.45\ M$

(c)  Mol fraction :-   Moles of water :-

$= \frac{80}{18} = 4.44$

So, mol fraction of KI :-

$= \frac{0.12}{0.12+4.44} = 0.0263$

### Question 2.6    $\inline H_{2}S,$ a toxic gas with rotten egg-like smell, is used for the qualitative analysis. If the solubility of $\inline H_{2}S,$  in water at STP is $\inline 0.195\; m.$ calculate Henry’s law constant.

For finding Henry's constant we need to know about the mole fraction of H2S.

Solubility of H2S in water is given to be 0.195 m .

i.e.,  0.195 moles in 1 Kg of water.

$Moles\: of\: water :=\frac{1000}{18} = 55.55\ moles$

So          $x_{H_2S} = Mole\ fraction\ of\ H_2S$

$= \frac{0.195}{0.195+55.55} = 0.0035$

At STP conditions, pressure = 1 atm or 0.987 bar

Equation is :                            $p_{H_2S} = K_h\times x_{H_2S}$

So we get :

$K_h =\frac{0.987}{0.0035} = 282\ bar$

### Question 2.7    Henry’s law constant for  $\inline CO_{2}$  in water is  $1.67\times 10^{8}\; Pa$ at  $\inline 298\; K.$ Calculate the quantity of  $\inline CO_{2}$  in $\inline 500\; mL$ of soda water when packed  under  $\inline 2.5\; atm$ $\inline CO_{2}$  pressure at $\inline 298 \; K.$

We know that ,

$p = k_h\times x$

Pressure of CO =  2.5 atm

We know that :              $1\ atm = 1.01\times10^5\ Pa$

So,   Pressure of CO2    =    $2.53\times10^5$ Pa

By Henry Law we get,

$x = \frac{p}{k_h} = \frac{2.53\times10^5}{1.67\times10^8} = 1.52\times10^{-3}$

Taking density of soda water = 1 g/ml

We get mass of water = 500 g.

So,                                Moles of water :

$= \frac{500}{18} = 27.78$

Also,                                      $x_{H_2O} = \frac{n_{CO_2}}{n_{H_2O}+n_{CO_2}} \approx \frac{n_{CO_2}}{n_{H_2O}}$

So,            moles of CO2  =  0.042 mol

Using relation of mole and given mass, we get

Mass of CO2  =   1.848 g.

## Question 2.8   The vapour pressure of pure liquids A and B are  $\inline 450$  and  $\inline 700\; mm \; Hg$ respectively, at $\inline 350\; K$ . Find out the composition of the liquid mixture if total vapour pressure is $\inline 600\; mm\; Hg$. Also find the composition of the vapour phase.

Let the composition of liquid A (mole fraction) be xA.

So mole fraction of B will be xB = 1 - xA.

Given that,    $P^{\circ}_A = 450\ mm\ of\ Hg\ ;\ P^{\circ}_B = 700\ mm\ of\ Hg$

Using Raoult’s law ,

$p_{total} = p^{\circ}_A\ x_A\ +\ p^{\circ} _B\ (1-x_A)$

Putting values of ptotal and vapour pressure of pure liquids in the above equation, we get :

600 =   450.xA  +    700.(1 -  xA)

or                           600 - 700  =  450xA - 700xA

or                                  xA   =   0.4

and                               xB    =   0.6

Now pressure in vapour phase :

$P_A = p^{\circ}_A\ x_A$

=   450(0.4)  = 180 mm of Hg

$P_B = p^{\circ}_B\ x_B$

=    700(0.6)   = 420 mm of Hg

$Mole\ fraction\ of\ liquid\ A = \frac{P_A}{P_A\ + P_B }$

$= \frac{180}{180\ + 420 } = 0.30$

And mole fraction of liquid B = 0.70

## Question 2.9    Vapour pressure of pure water at $298 \; K$ is $\inline 23.8\; mm \; Hg.$ $\inline 50\; g$ of urea $\inline (NH_{2}CONH_{2})$ is dissolved in $\inline 850 \; g$ of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Given that vapour pressure of pure water,  $p^{\circ}_w = 23.8\ mm\ of\ Hg$

Moles of water  :

$= \frac{850}{18} = 47.22$

Moles of urea  :

$= \frac{50}{60} = 0.83$

Let the vapour pressure of water be pw.

By Raoult's law, we get :

$\frac{p^{\circ}_w - p_w}{p^{\circ}_w} = \frac{n_2}{n_1\ + n_2}$

or                         $\frac{23.8 - p_w}{23.8} = \frac{0.83}{47.22\ + 0.83}=0.0173$

or                          pw  =   23.4 mm of Hg.

Relative lowering :-  Hence, the vapour pressure(v.P) of water in the solution = 23.4 mm of Hg

and its relative lowering = 0.0173.

### Question 2.10    Boiling point of water at $\inline 750 \; mm \; Hg$  is  $\inline 99.63^{\circ}C$. How much sucrose is to be added to $\inline 500 \; g$ of water such that it boils at $\inline 100^{\circ}C.$

Here we will use the formula :

$\Delta T_b = \frac{K_b\times1000\times w_2}{M_2\times w_1}$

Elevation in temperature =  100 -  99.63  =  0.37

Kb = 0.52 ;     $Molar\ mass\ of\ sucrose = 11(12) + 22(1) + 11(16) = 342\ g\ mol^{-1}$

Putting all values in above formula, we get :

$w_2 = \frac{0.37\times342\times500}{0.52\times1000}= 121.67\ g$

Thus 121.67 g of sucrose needs to be added.

## Question 2.11  Calculate the mass of ascorbic acid (Vitamin C, $\inline C_{6}H_{8}O_{6}$ ) to be dissolved in $\inline 75 \; g$ of acetic acid to lower its melting point by $\inline 1.5^{\circ}C$.           $\inline K_{f}=3.9\; K\; kg\; mol^{-1}$

Elevation in melting point = 1.5 degree celsius.

Here we will use the following equation :

$\Delta T_b = \frac{K_b\times 1000\times w_2}{M_2\times w_1}$

Putting given values in the above equation :

$w_2 = \frac{1.5\times176\times75}{3.9\times1000} = 5.08\ g$

Thus 5.08 ascorbic acid is needed for required condition.

### Question 2.12   Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving $1.0 \; g$ of polymer of molar mass $\inline 185,000$  in $\inline 450\; mL$ of water at $\inline 37^{\circ}C.$

We know that :

$Osmotic\ Pressure = \Pi = \frac{n}{v}RT$

We are given with  :-

$Moles\ of\ polymer = \frac{1}{185000}$

Volume, V = 0.45 L

Thus osmotic pressure :

$= \frac{\frac{1}{185000}\times8.314\times10^3\times310}{0.45} = 30.98\ Pa$

NCERT solutions for class 12 chemistry chapter 2 Solutions : Exercises

### Q.2.1 Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

Solution :-  A solution is a homogeneous mixture of two or more non-reacting substances. It has two components :- solute and solvent.

Types of solutions are given below :-

### Q.2.2 Give an example of a solid solution in which the solute is a gas.

Solution of hydrogen in palladium is such an example in which solute is a gas and solvent is solid.

### Q.2.3(i) Define the following terms:

Mole fraction

Mole fraction is defined as the ratio of number of moles of a component and total number of moles in all components.

i.e.,                          $Mole\ fraction = \frac{Number\ of\ moles\ in\ a\ component }{Total\ number\ of\ moles\ in\ all\ components}$

### Q.2.3(ii) Define the following terms:

Molality

It is defined as the number of moles of solute dissolved per kg (1000g) of solvent

i.e.,                              $Molality = \frac{Number\ of\ moles\ of\ solute}{Mass\ of\ solvent\ in\ Kg}$

It is independent of temperature.

## Q2.3(iii) Define the following terms:

Molarity

Molarity is defined as number of moles of solute dissolved per litre(or 1000ml)  of solution.

i.e.,                                     $Molarity = \frac{No.\ of\ moles\ of\ solute}{Volume\ of\ solution\ in\ litre}$

It depends on temperature because volume is dependent on temperature.

## Q2.3(iv) Define the following terms:

Mass percentage.

Mass percentage is defined as the percentage ratio of mass of one component to the total mass of all the components.

i.e.,                       $Mass\ percentage = \frac{Mass\ of\ a\ component}{Total\ mass\ of\ solution}\times100$

## Q2.4 Concentrated nitric acid used in laboratory work is $68\%$ nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is  $\inline 1.504\; g\; mL^{-1}\; ?$

According to given question, in 100 g of solution 68 g is nitric acid and rest is water.

So moles of 68 g HNO3 :-

$=\ \frac{68}{63} = 1.08$

Density of solution is given to be 1.504.

So volume of 100 g solution becomes :-

$=\ \frac{100}{1.504} = 66.49\ mL$

Thus, molarity of nitric acid is :

$Molarity = \frac{1.08}{\frac{66.49}{1000}} = 16.24\ M$

## Q2.5 A solution of glucose in water is labelled as $10\%$  w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is $\inline 1.2\; g\; mL^{-1},$  then what shall be the molarity of the solution?

According to question, $10\%$ mass percentage means in 100 g of solution 10 g glucose is dissolved in 90 g water.

Molar mass of glucose (C6H12O6)  =  $180\ g\ mol^{-1}$

So moles of glucose are :

$\frac{10}{180} = 0.056 mol$

$Moles\: of\: water = \frac{90}{18} = 5 mol$

$Molality = \frac{0.056}{0.09} = 0.62\ m$

Mole fraction :-

$\frac{0.056}{0.056+5} = 0.011$

Molarity :-             Volume of 100 g solution  :

$=\frac{100}{1.2} =83.3\ mL$

$Molarity = \frac{0.056}{83.33\times10^{-3}} =0.67\ M$

### Q2.6 How many mL of $\inline 0.1\; M\; HCl$ are required to react completely with $\inline 1 \; g$ mixture of $\inline Na_{2}CO_{3}$  and $\inline NaHCO_{3}$  containing equimolar amounts of both?

Total amount of mixture of Na2CO3 and NaHCO3 = 1 g.

Let the amount of Na2CO3 be x g.

So the amount of NaHCO will be equal to (1 - x) g.

$Molar\ mass\ of\ Na_2CO_3 = 106\ ;\ molar\ mass\ of\ NaHCO_3 = 84$

Now it is given that it is an equimolar mixture.

So,            Moles of Na2CO3 =  Moles of  NaHCO3.

or                          $\frac{x}{106} = \frac{1-x}{84}$

or                            x = 0.558 g

So                       $Moles\ of \ Na_2CO_3 = \frac{0.558}{106} = 0.00526$

and                     $Moles\ of \ NaHCO_3 = \frac{1 - 0.558}{84} = 0.0053$

It is clear that for 1 mol of Na2CO3  2 mol of HCl is required, similarly for 1 mol of NaHCO3 1 mol of HCl is required.

So number of moles required of HCl =  2(0.00526) + 0.0053 = 0.01578 mol

It is given that molarity of HCl is 0.1 which means 0.1 mol of HCl in 1l of solution.

Thus required volume :

$= \frac{0.01578}{0.1} = 0.1578\ l = 157.8\ mL$

### Q2.7 A solution is obtained by mixing  $\inline 300 \; g$ of $25\%$ solution and $\inline 400 \; g$ of  $40\%$  solution by mass. Calculate the mass percentage of the resulting solution.

According to question we have 2 solute,

Solute 1. :  $25\%$ of 300 g gives :

$\frac{25}{100}\times300 = 75\ g$

Solute 2.  :  $40\%$ of 400 g gives :

$\frac{40}{100}\times400 = 160\ g$

So total amount of solute =  75 + 160 = 235 g.

Thus mass percentage of solute is :

$= \frac{235}{700}\times100 = 33.5\%$

and mass percentage of water   $= 100 - 33.5 = 66.5\%$

## Q2.8 An antifreeze solution is prepared from $\inline 222.6\; g$ of ethylene glycol  $\inline (C_{2}H_{6}O_{2})$  and $\inline 200\; g$ of water. Calculate the molality of the solution. If the density of the solution is $\inline 1.072 \; g\; mL^{-1}$ then what shall be the molarity of the solution?

For finding molality we need to find the moles of ethylene glycol.

Moles of ethylene glycol :

$= \frac{222.6}{62}= 3.59\ mol$

We know that :

$Molality = \frac{Moles\ of\ ethylene\ glycol}{Mass\ of\ water}\times100$

$= \frac{3.59}{200}\times100 = 17.95\ m$

Now for molarity :-

Total mass of solution = 200 + 222.6 = 422.6 g

Volume of solution

$= \frac{422.6}{1.072}= 394.22\ mL$

So molarity :-

$= \frac{3.59}{394.22}\times1000= 9.11\ M$

### Q2.9(i) A sample of drinking water was found to be severely contaminated with chloroform $\inline (CHCl_{3})$ supposed to be a carcinogen. The level of contamination was $\inline 15\; ppm$ (by mass):

express this in percent by mass

We know that 15 ppm means 15 parts per million.

Required percent by mass :

$= \frac{Mass\ of\ chlorofoam}{Total\ mass}\times100$

$= \frac{15}{10^6}\times100 = 1.5\times10^{-3}\%$

## Q2.9(ii) A sample of drinking water was found to be severely contaminated with chloroform $\inline (CHCl_{3})$  supposed to be a carcinogen. The level of contamination was $\inline 15\; ppm$ (by mass):

determine the molality of chloroform in the water sample.

Moles of chloroform  :

$=\frac{15}{119.5} = 0.1255\ mol$

Mass of water is $10^6$.      (Since contamination is 15 ppm)

So molality will be :

$=\frac{0.1255}{10^6}\times1000 = 1.255\times10^{-4}\ m$

### Q2.10 What role does the molecular interaction play in a solution of alcohol and water?

Both alcohol and water individually have strong hydrogen bonds as their force of attraction. When we mix alcohol with water they form solution due to the formation of hydrogen bonds but they are weaker as compared to hydrogen bonds of pure water or pure alcohol.

Thus this solution shows a positive deviation from the ideal behaviour.

### Q2.11 Why do gases always tend to be less soluble in liquids as the temperature is raised?

It is known that dissolution of gas in a liquid is an exothermic process. So, by Le Chatelier principle we know that equilibrium shifts backwards as we increase temperature in case of exothermic process. Thus gases always tend to be less soluble in liquids as the temperature is raised.

According to Henry's law at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of solution or liquid.

i.e.,   p = khx, Here kh is Henry’s law constant.

Some of its applications are as follows:-

(a)  We can increase the solubility of CO2 in soft drinks, the bottle is sealed under high pressure.

(b)  To avoid bends (due to blockage of capillaries) and the toxic effects of high concentrations of nitrogen in the blood, the tanks used by scuba divers are filled with air diluted with helium (11.7% helium, 56.2% nitrogen and 32.1% oxygen).

(c)  The partial pressure of oxygen is less at high altitudes than that at the ground level. This leads to low concentrations of oxygen in the blood and tissues climbers. Due to low blood oxygen, climbers become weak and unable to think clearly which are symptoms of a condition known as anoxia.

Using Henry's Law we can write,

$m = k.P$

Putting value in this equation, we get :

$6.56\times10^{-3} = k\times 1$

So, the magnitude of k is   $6.56\times10^{-3}$.

Now, we will again use the above equation for  $m = 5.0\times10^{-2}\ g$.

So the required partial pressure is :-

$p = \frac{m}{k} = \frac{5.0\times10^{-2}}{6.56\times10^{-3}}$

or                                                    $p = 7.62\ bar$

Positive and negative deviation:- A non-ideal solution is defined as a solution which does not obey Raoult’s law over the entire range of concentration i.e.,     $\Delta _{Mix}H \neq 0$ and $\Delta _{Mix}V \neq 0$. The vapour pressure of these solutions is either higher or lower than that expected by Raoult’s law. If vapour pressure is higher, the solution shows a positive deviation and if it is lower, it shows a negative deviation from Raoult’s law.

Enthalpy relation to positive and negative deviation can be understood from the following example:-

Consider a solution made up of two components -  A and B. In the pure state the intermolecular force of attraction between them are A-A and B-B. But when we mix the two, we get a binary solution with molecular interaction A-B.

If A-B interaction is weak than A-A and B-B then enthalpy of reaction will be positive thus reaction will tend to move in a backward direction. Hence molecules in binary solution will have a higher tendency to escape. Thus vapour pressure increases and shows positive deviation from the ideal behaviour.

Similarly, for negative deviation, A-B interaction is stronger than that of A-A and B-B.

It is given that $2\%$ of aq. solution. This means 2 g of non-volatile solute in 98 g of H2O.

Also the vapour of water at normal boiling point = 1.013 bar.

Using Raoult's law :

$\frac{p_w^{\circ} - p}{p_w^{\circ}} = \frac{w_2.M_1}{w_1.M_2}$

So we get :

$M_2 = \frac{2\times18\times1.013}{0.009\times98} = 41.35\ g\ mol^{-1}$

Thus the molar mass of non-volatile solute is 41.35 unit.

Vapour pressure of heptane =   $p_h^{\circ} = 105.2\ KPa$

and vapour pressure of octane =  $p_o^{\circ} = 46.8\ KPa$

Firstly we will find moles of heptane and octane so that we can find vapour pressure of each.

Molar mass of heptane =  7(12) + 16(1)  =  100 unit.

and molar mass of octane =  8(12) + 18(1) = 114 unit.

So moles of heptane :

$\frac{26}{100} = 0.26$

and moles of octane :

$\frac{35}{114} = 0.31$

Mole fraction of heptane =  0.456 and mole fraction of octane = 0.544

Now we will find the partial vapour pressure:-

(i) of heptane :-  $p_h = 0.456\times105.2 = 47.97\ KPa$

(ii) of octane  :-   $p_o = 0.544\times46.8 = 25.46\ KPa$

So total pressure of solution = $p_h+p_o$

=  47.97 + 25.46 =   73.43 KPa

It is asked the vapour pressure of 1 molal solution which means 1 mol of solute in 1000 g H2O.

Moles in 1000g of water = 55.55 mol.         (Since the molecular weight of H2O is 18)

Mole fraction of solute :

$\frac{1}{1+55.55} = 0.0177$

Applying the equation :

$\frac{p_w^{\circ} - p}{p_w^{\circ}} = x_2$

or                                       $\frac{12.3 - p}{12.3} = 0.0177$

or                                            $p = 12.083\ KPa$

Thus the vapour pressure of the solution is 12.083 KPa

Let the initial vapour pressure of octane = $p_o^{\circ}$.

After adding solute to octane, the vapour pressure becomes :

$=\frac{80}{100}\times p_o^{\circ} = 0.8p_o^{\circ}$

Moles of octane :

$= \frac{114}{114} = 1$               $\left ( Molar\ mass\ of\ octane = 8(12) + 18(1) = 114\ g\ mol^{-1} \right )$

Using Raoult's law we get :

$\frac{p_o^{\circ} - p}{p_o^{\circ}} = x_2$

or                                     $\frac{p_o^{\circ} -0.8p_o^{\circ} }{p_o^{\circ}} = \frac{\frac{W}{40}}{\frac{W}{40}+1}$

or                                            $w = 10\ g$

Thus required mass of non-volatile solute = 10g.

## Q2.19(i)  A solution containing 30 g of non-volatile solute exactly in $\inline 90\; g$ of water has a vapour pressure of $\inline 2.8\; k\; Pa$ at $\inline 298 \; K.$. Further, $\inline 18 \; g$ of water is then added to the solution and the new vapour pressure becomes $\inline 2.9\; kPa$ at $\inline 298\; K.$. Calculate:

molar mass of the solute

In this question we will find molar mass of solute by using Raoult's law .

Let the molar mass of solute is M.

Initially we have 30 g solute and 90 g water.

Moles of water :

$\frac{90}{18} = 5\ mol$

By Raoult's law we have :-

$\frac{p_w^{\circ} - p}{p_w^{\circ}} = \frac{n_2}{n_1+n_2}$

or                                        $\frac{p_w^{\circ} - 2.8}{p_w^{\circ}} = \frac{\frac{30}{M}}{5+\frac{30}{M}} = \frac{30}{5M+30}$

or                                         $\frac{p_w^{\circ} }{2.8} = \frac{5M+30}{5M}$                                                                                ------------------------------   (i)

Now we have added 18 g of water more, so the equation becomes:

Moles of H2O  :

$\frac{90+18}{18} = 6\ mol$

Putting this in above equation we obtain :-

$\frac{p_w^{\circ} - 2.9}{p_w^{\circ}} = \frac{\frac{30}{M}}{6+\frac{30}{M}} = \frac{30}{6M+30}$

or                                  $\frac{p_w^{\circ} }{2.9} = \frac{6M +30}{6M}$                                                                                   -----------------------------------(ii)

From equation (i) and (ii) we get

M  =  23 u

So the molar mass of solute is 23 units.

## Q2.19(ii) A solution containing $\inline 30\; g$ of non-volatile solute exactly in $\inline 90\; g$ of water has a vapour pressure of  $\inline 2.8\; kPa$ at $\inline 298 \; K$ Further, $\inline 18 \; g$ of water is then added to the solution and the new vapour pressure becomes $\inline 2.9 \; kPa$ at $\inline 298 \; K.$ Calculate

vapour pressure of water at 298 K.

In the previous part we have calculated the value of molar mass the Raoul's law equation.

$\frac{p_w^{\circ} }{2.8} = \frac{5M+30}{5M}$

Putting M = 23 u in the above equation we get,

$\frac{p_w^{\circ} }{2.8} = \frac{5(23)+30}{5(23) } =\frac{145}{115}$

or                         $p_w^{\circ} =3.53$

Thus vapour pressure of water = 3.53 kPa.

## Q2.20 A  $5\%$ solution (by mass) of cane sugar in water has freezing point of  $\inline 271 K.$  Calculate the freezing point of  $5\%$  glucose in water if freezing point of pure water is $\inline 273.15 \; K.$

It is given that freezing point of pure water is 273.15 K.

So, elevation of freezing point = 273.15 - 271 = 2.15 K

$5\%$ solution means 5 g solute in 95 g of water.

Moles of cane sugar :

$= \frac{5}{342} = 0.0146$

Molality :

$= \frac{0.0146}{0.095} = 0.1537$

We also know that -              $\Delta T_f = k_f \times m$

or                                          $k_f = \frac{\Delta T_f}{m} = 13.99\ K\ Kg\ mol^{-1}$

Now we will use the above procedure for glucose.

$5\%$ of glucose means 5 g of gluocse in 95 g of H2O.

Moles of glucose :

$\frac{5}{180} = 0.0278$

Thus molality :

$= \frac{.0278}{0.095} = 0.2926\ mol\ kg^{-1}$

So,  we can find the elevation in freezing point:

$\Delta T_f = k_f \times m$

$= 13.99 \times 0.2926 = 4.09\ K$

Thus freezing point of glucose solution is 273.15 - 4.09  = 269.06 K.

## Q2.21 Two elements A and B form compounds having formula AB2 and AB4 . When dissolved in 20 g of benzene $\inline (C_{6}H_{6}),$ $\inline 1\; g$ of AB2 lowers the freezing point by $\inline 2.3\; K$ whereas $\inline 1.0\; g$ of AB4 lowers it by $\inline 1.3\; K$. The molar depression constant for benzene is $\inline 5.1\; K\; kg\; mol^{-1}.$Calculate atomic masses of A and B.

In this question we will use the formula :

$\Delta T_f = k_f \times m$

Firstly for compound AB2  :-

$M_B = \frac{K_f\times W_b \times 1000}{w_A\times \Delta T_f}$

or                                                       $= \frac{5.1\times 1 \times 1000}{20\times 2.3} = 110.87\ g/mol$

Similarly for compound AB4 :-

$M_B= \frac{5.1\times 1 \times 1000}{20\times 1.3} = 196\ g/mol$

If we assume atomic weight of element A to be x and of element B to be y, then we have :-

x + 2y = 110.87                               -----------------   (i)

x + 4y = 196                                    -----------------   (ii)

Solving both the equations, we get :-

x  =  25.59  ;    y =  42.6

Hence atomic mass of element A is 25.59u and atomic mass of element B is 42.6u.

### Q2.22 At $\inline 300 \; K,$ 36 g of glucose present in a litre of its solution has an osmotic pressure of $\inline 4.98\; bar.$ If the osmotic pressure of the solution is $\inline 1.52\; bars$ at the same temperature, what would be its concentration?

According to given conditions we have same solution under same temperature. So we can write :

$\frac{\Pi _1}{C_1} = \frac{\Pi _2}{C_2}$                                                   $\left ( \Pi = C.R.T\ ; RT = \frac{\Pi }{C} \right )$

So, if we put all the given values in above equation, we get

$\frac{4.98}{\frac{36}{180}} = \frac{1.52}{C_2}$

or                                              $C_2 = \frac{1.52\times36}{4.98\times180} = 0.061 M$

Hence the required concentration is 0.061 M.

### Q2.23(i) Suggest the most important type of intermolecular attractive interaction in the following pairs.

n-hexane and n-octane

Since both the compounds are alkanes so their mixture has van der Waal force of attraction between compounds.

### Q2.23(ii) Suggest the most important type of intermolecular attractive interaction in the following pairs.

$\inline I_{2}\; and\; CCl_{4}$

The binary mixture of these compounds has van der Waal force of attraction between them.

## $\inline NaClO_{4} \; and \; Water$

The given compounds will have ion-dipole interaction between them.

### Q2.23(iv) Suggest the most important type of intermolecular attractive interaction in the following pairs.

$methanol\; and\; acetone$

Methanol has -OH group and acetone has ketone group. So there will be hydrogen bonding between them.

### Q2.23(v) Suggest the most important type of intermolecular attractive interaction in the following pairs.

acetonitrile $\inline (CH_{3}CN)$ and acetone $\inline (C_{3}H_{6}O).$

They will have dipole-dipole interaction since both are polar compounds.

### Q2.24 Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain.

Cyclohexane, KCl, CH3OH, CH3CN.

The order will  be :           Cyclohexane > CH3CN > CH3OH > KCl

In this, we have used the fact that like dissolves like.

Since cyclohexane is an alkane so its solubility will be maximum.

## phenol

We know the fact that like dissolves in like.

Since phenol is had both polar and non-polar group so it is partially soluble in water.

### Q2.25(ii)  Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?

toluene

Since toluene is a non-polar compound i.e., it doesn't have any polar group so it is insoluble in water. (because water is a polar compound)

## formic acid

Since the -OH group in formic acid (polar) can form H-bonds with water thus it is highly soluble in water.

### Q2.25(iv)  Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?

ethylene glycol

Ethylene glycol is an organic compound but is polar in nature. Also, it can form H-bonds with water molecules, thus it is highly soluble in water.

### Q2.25(v) Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?

chloroform

Chloroform is a non-polar compound so it is insoluble in water.

## Q2.25(vi) Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?

pentanol

Pentanol has both polar and non-polar groups so it is partially soluble in water.

## Q2.26 If the density of some lake water is $\inline 1.25g\; mL^{-1}$ and contains $\inline 92\; g$ of $\inline Na^{+}\; ions$ per kg of water, calculate the molality of $\inline Na^{+}\; ions$ in the lake.

We know that, Molality :

$Molality = \frac{Moles\ of\ solute}{Mass\ of\ solvent\ in\ Kg}$

So, for moles of solute we have :

$Moles\ of\ Na^+ = \frac{92}{23} = 4$

Thus,  molality :

$= \frac{4}{1} = 4$

Molality of Na+ ions is 4m.

### Q2.27  If the solubility product of $\inline CuS$ is $6\times10^{-16}$, calculate the maximum molarity of $\inline CuS$ in aqueous solution.

We are given,    $K_{sp} = 6\times10^{-16}$

The dissociation equation of CuS is given by :-

So, the equation becomes :-   $K_{sp} = Cu^{2+}\times {S^{2-}}$

or                                            $K_{sp} = s\times s = s^2$

or                                                     $s = 2.45\times 10^{-8}\ M$

Thus maximum molarity of solution is   $2.45\times 10^{-8}\ M$.

### Q2.28 Calculate the mass percentage of aspirin  $\inline (C_{9}H_{8}O_{4})$  in acetonitrile $\inline (CH_{3}CN)$ when $\inline 6.5 \; g$ of $\inline C_{9}H_{8}O_{4}$ is dissolved in $\inline 450\; g$ of $\inline CH_{3}CN$.

Total mass of solution = Mass of aspirin + Mass of acetonitrile = 6.5 + 450 = 456.5 g.

We know that :

$Mass\ percentage = \frac{Mass\ of\ solute }{Mass\ of\ solution}\times100$

So,                      $Mass\ percentage = \frac{6.5 }{456.5}\times100 = 1.42\%$

Thus the mass percentage of aspirin is$1.42\%$

### Q2.29 Nalorphene $\inline (C_{19}H_{21}NO_{3})$, similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is $\inline 1.5 \; mg.$ Calculate the mass of $\inline 1.5 -10^{-3} m$ aqueous solution required for the above dose.

We are given with molality of the solution, so we need to find the moles of Nalorphene.

Molar mass of nalorphene =  19(12) + 21(1)  + 1(14) + 3(16)  =  311u.

So moles of nalorphene :

$\frac{1.5\times10^{-3}}{311} = 4.82\times 10^{-6}\ moles$

Molality :

$= \frac{No.\ of\ moles\ of\ solute}{Mass\ of\ solvent\ in\ kg}$

or                                                       $1.5\times 10^{-3} = \frac{4.82\times 10^{-6}}{w}$

or                                                               $w = 3.2\times 10^{-3}\ Kg$

So the required weight of water is 3.2 g.

### Q2.30 Calculate the amount of benzoic acid $\inline (C_{6}H_{5}COOH)$ required for preparing $\inline 250\; mL$  of  $\inline 0.15 \; M$ solution in methanol.

Molar mass of benzoic acid = 7(12) + 6(1) + 2(16) = 122u.

We are given with the molarity of solution.

$Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution\ in\ litre}$

or                                                          $0.15 = \frac{Moles\ of\ solute}{\frac{250}{1000}}$

or                                      $Moles\ of\ solute= \frac{0.15\times 250}{1000} = 0.0375\ mol$

So mass of benzoic acid :

$= Moles\ of\ benzoic\ acid \times Molar\ mass$

$=0.0375\times 122 = 4.575\ g$

Hence the required amount of benzoic acid is 4.575 g.

### Q2.31 The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.

We know that depression in freezing point of water will depend upon the degree of ionisation.

The degree of ionisation will be highest in the case of trifluoroacetic acid as it is most acidic among all three.

The order of degree of ionisation on the basis if acidic nature will be:-     Trifluoroacetic acid > Trichloroacetic acid > Acetic acid.

So the depression in freezing point will be reverse of the above order.

### Q2.32 Calculate the depression in the freezing point of water when $\inline 10 \; g$ of  $\inline CH_{3}CH_{2}CHCICOOH$  is added to $\inline 250 \; g$  of water.

Firstly we will find the Vant's Hoff factor the dissociation of given compound.

So we can write,                                             $K_a = \frac{(Ca\times Ca)}{C(1-a)}$

or                                                                     $K_a = Ca^2$                                                  $(\because a << 1)$

or                                                                      $a = \sqrt{\frac{K_a}{C}}$

Putting values of Ka and C in the last result, we get :

$a = 0.0655$

At equilibrium   i =  1 - a + a + a = 1 + a = 1.0655

Now we need to find the moles of the given compound CH3CH2CHClCOOH.

So, moles =

$\frac{10}{122.5} = 0.0816\ mol$

Thus, molality of the solution :

$= \frac{0.0816\times1000}{250} = 0.3265\ m$

Now we will use :

$\Delta T_f = i\ K_f\ m$

or                                           $= 1.065 \times 1.86 \times 0.3265 = 0.6467\ K$

### Q2.33 $\inline 19.5 \; g$ of $\inline CH_{2}FCOOH$ is dissolved in $\inline 500\; g$  of water. The depression in the freezing point of water observed is $\inline 1.0^{\circ}C$. Calculate the van’t  Hoff factor and dissociation constant of fluoroacetic acid.

Firstly we need to calculate molality in order to get vant's hoff factor.

So moles of CH2FCOOH :

$\frac{19.5}{78} = 0.25$

We need to assume volume of solution to be nearly equal to 500 mL.         (as 500 g water is present)

Now, we know that :            $\Delta T_f = i\ K_f\ m$

or                                           $i = \frac{1}{0.93} = 1.0753$

Now for dissociation constant :-

$a = i - 1 = 1.0753 - 1 = 0.0753$

and,                                            $K_a = \frac{Ca^2}{1-a}$

Put values of C and a in the above equation, we get :

$K_a = 3\times10^{-3}$

## Q2.34 Vapour pressure of water at  $\inline 293 \; K$ is  $\inline 17.535\; mm\; Hg.$ Calculate the vapour pressure of water at $\inline 293 \; K$  when  $\inline 25 \; g$  of glucose is dissolved in $\inline 450 \; g$ of water.

Firstly we will find number of moles of both water and glucose.

Moles of glucose :

$= \frac{25}{180} = 0.139\ mol$                                        $(Molar\ mass\ of\ glucose = 6(12)+12(1)+6(16) = 180\ g\ mol^{-1})$

and moles of water :

$= \frac{450}{18} = 25\ mol$                                            $(Molar\ mass\ of\ water = 18\ g\ mol^{-1})$

Now,

$\frac{p_w^{\circ} - p}{p_w^{\circ}} = \frac{n_g}{n_g + n_w}$

or                                        $\frac{17.535 - p}{17.535} = \frac{0.139}{0.139+ 25}$

or                                          $p = 17.44\ mm\ of\ Hg$

Thus vapour pressure of water after glucose addition = 17.44 mm of Hg

### Q2.35 Henry’s law constant for the molality of methane in benzene at $\inline 298\; K$ is $4.27\times10^{5} mm\ Hg$. Calculate the solubility of methane in benzene at  $\inline 298\; K$ under $\inline 760\; mm\; Hg.$

We know that :            $P = k\times C$

We are given value of P and k, so C can be found.

$C = \frac{760}{4.27\times10^5} = 178\times 10^{-5}$

Hence solubility of methane in benzene is $178\times 10^{-5}$.

### Q2.36 $\inline 100\; g$ of liquid A $\inline (molar\; mass \; 140\; g\; mol^{-1})$ was dissolved in $\inline 1000\; g$ of liquid B $\inline (molar\; mass\; 180\; g\; mol^{-1})$. The vapour pressure of pure liquid B was found to be $\inline 500 \; Torr.$Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure the solution is  $\inline 475 \; Torr.$

For calculating partial vapour pressure we need to calculate mole fractions of components.

So number of moles of liquid A :

$= \frac{100}{140} = 0.714$

and  moles of liquid B :

$= \frac{1000}{180} = 5.556$

Mole fraction of A (xA)   :

$= \frac{0.714}{0.714+5.556} = 0.114$

and mole fraction of B (xB)  :

$= \frac{5.556}{0.714+5.556} = 0.866$

Now,                    Ptotal  =  PA + PB

or                         $P_{total} = P_A^{\circ}x_A\ + P_B^{\circ}x_B$

or                          $475 = P_A^{\circ}\times0.114\ + 500\times0.886$

or                            $P_A^{\circ} = 280.7\ torr$

Thus vapour pressure in solution due to A =   $P_A^{\circ}x_A$

$= 280.7 \times0.114 = 32\ torr$

### Q2.37 Vapour pressures of pure acetone and chloroform at $\inline 328 \; K$ are $\inline 741.8 \; mm\; Hg$ and $\inline 638.8 \; mm\; Hg$  respectively. Assuming that they form ideal solution over the entire range of composition, plot ptotal, pchloroform, and pacetone as a function of xacetone. The experimental data observed for different compositions of mixture is:

 $\inline 100 \times x_{acetone}$                            0         11.8        23.4      36.0     50.8     58.2     64.5      72.1 $\inline p_{acetone}/mm\; Hg$                       0         54.9      110.1     202.4    322.7   405.9   454.1   521.1 $\inline p_{chloroform}/mm\; Hg$             632.8     548.1    469.4      359.7    257.7   193.6   161.2   120.7 $p_{total}$                                        632.8    603         579.5     562.1    580.4   599.5   615.5    64.18

Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.