NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry

 

NCERT solutions for class 11 chemistry chapter 1 Some Basic Concepts of Chemistry - Chapter 1 is one of the important chapters and consists of important concepts that are often asked in class 11 final examination. In this chapter, there are 36 questions in the exercise. The NCERT solutions for class 11 chemistry chapter 1 Some Basic Concepts of Chemistry are designed systematically by the chemistry subject experts. This NCERT solutions help students in their preparation of class 11 final examination and also helps in competitive exams like JEE, NEET, BITSAT, etc.

When we start studying chemistry, different kinds of questions comes to our mind like what is chemistry, why we are going to study in chemistry and what is the importance of chemistry in our life? To solve these type of questions we require some basic concepts of chemistry and basic techniques of chemistry. In solutions of NCERT for class 11 chemistry chapter 1 Some Basic Concepts of Chemistry, we will discuss questions based on only such concepts and techniques. In this chapter, we deal with topics like the importance of chemistry, different states of matter, atomic and molecular mass, Dalton’s atomic theory, Avogadro Law, the law of conservation of mass, normality, etc. After studying CBSE NCERT solutions for class 11 chemistry chapter 1 Some Basic Concepts of Chemistry, students will be able to understand the role of chemistry in our daily life; explain the characteristics of three states of matter which are solid, liquid and gas; classify different substances into elements, mixtures and compounds; able to explain various laws of chemical combination; describe terms like mole, molar mass, mole fraction, molarity and molality and also determine molecular formula and empirical formula. Please scroll down below to get all the NCERT solutions for class 11 chemistry chapter 1 Some Basic Concepts of Chemistry.

 

Formulas of Chapter 1 Some Basic Concepts of Chemistry

1. The mass percent of an element

    \\Mass\:percent\:of =\frac{Mass \:of\:that\:element\:in \:the\:compound\times 100}{molar \:mass\:of\:the\:compound}\\an\:element

2. Mass percent 

    \\Mass\:percent =\frac{Mass \:of\:solute\times 100}{mass\:of\:solution}\\

3. Mole fraction

   \\Mole\: fraction \:of \:A =\frac{No. \:of\:moles\:of\:A}{No.\:of\:moles\:of\:solutions}=\frac{n_A}{n_A+n_B}

4. Molarity(M)

   \\M =\frac{No. \:of\:moles\:of\:solute}{Volume\:of\:solution\:in\:litres}

5. Molality(m)

   \\m =\frac{No. \:of\:moles\:of\:solute}{Mass\:of\:solvent\:in\:Kg}

Topics of NCERT Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry

1.1 Importance of Chemistry

1.2 Nature of Matter

1.3 Properties of Matter and their Measurement

1.4 Uncertainty in Measurement

1.5 Laws of Chemical Combinations

1.6 Dalton’s Atomic Theory

1.7 Atomic and Molecular Masses

1.8 Mole Concept and Molar Masses

1.9 Percentage Composition

1.10 Stoichiometry and Stoichiometric Calculations

 

NCERT solutions for class 11 chemistry chapter 1 Some Basic Concepts of Chemistry - Exercise Questions

 

Question 1.1    Calculate the molar mass of the following:
                (i) H2

Answer:

The molar mass of Water (H_{2}O) is:

= 2\times Atomic\ mass\ of\ hydrogen\ + \ Atomic\ mass\ of\ oxygen

= 2\times(1.008\ u) + 16.00\ u

= 18.02\ u

Question 1.1    Calculate the molar mass of the following:
                (ii) CO2 

Answer:

The molar mass of Carbon dioxide CO_{2} is:

= 1\times Atomic\ mass\ of\ carbon\ + \ 2\times Atomic\ mass\ of\ oxygen

= 1\times(12.011\ u) + 2\times (16.00\ u)

= 44.011\ u

 

 

Question 1.1   Calculate the molar mass of the following:
                (iii) CH 4

Answer:

The molar mass of Methane CH_{4} is:

= 1\times Atomic\ mass\ of\ carbon\ + \ 4\times Atomic\ mass\ of\ hydrogen

= 1\times(12.011\ u) + 4\times (1.008\ u)

= 16.043\ u

 

Question 1.2    Calculate the mass per cent of different elements present in sodium sulphate (Na2SO4 ).

Answer:

The different elements present in sodium sulphate are: 

Sodium (Na), Sulphur (S), and Oxygen (O).

The molar mass of sodium sulphate Na_{2}SO_{4};= (2\times Atomic\ mass\ of\ Na) + (1\times Atomic\ mass\ of\ S) + (4\times Atomic\ mass\ of\ O)

= 2\times (23.0\ u) + (32.066\ u) + 4\times (16.00\ u)

= 142.066\ u

So, Mass percentage of an element in a compound is given by,

= \frac{Mass\ of\ element\ in\ the\ compound}{Molar\ mass\ of\ the\ compound} \times 100

Therefore,

The mass per cent of Sodium (Na):

=\frac{46.0\ u}{142.066\ u} \times 100

=32.379\%

\approx 32.4\%

 The mass per cent of Sulphur (S):

=\frac{32.066\ u}{142.066\ u} \times 100

= 22.57 \%

\approx 22.6\%

The mass per cent of Oxygen (O):

=\frac{64.0\ u}{142.066\ u} \times 100

= 45.049 \%

\approx 45.0\%

 

 

Question 1.3    Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.

Answer:

Given there is an oxide of iron which has 69.9\% iron and 30.1\% dioxygen by mass:

Relative moles of iron in iron oxide:

= \frac{\%\ of\ iron\ by\ mass}{Atomic\ mass\ of\ iron}

= \frac{69.9}{55.85} = 1.25

Relative moles of oxygen in iron oxide:

= \frac{\%\ of\ oxygen\ by\ mass}{Atomic\ mass\ of\ oxygen}

= \frac{30.1}{16.00} = 1.88

The simplest molar ratio of iron to oxygen:

\Rightarrow 1.25:1.88 \Rightarrow 1:1.5 \Rightarrow 2:3

Therefore, the empirical formula of the iron oxide is Fe_{2}O_{3}.

 

Question 1.4    Calculate the amount of carbon dioxide that could be produced when

                (i) 1 mole of carbon is burnt in air.

Answer:

When carbon is burnt in the air:

The chemical equation for this reaction is:

C_{(s)}+O_{2(g)} \rightarrow CO_{2(g)}

Here 1 mole of carbon (solid) weighing 12g is burnt in 1 mole of Dioxygen (gas) weighing 32g to produced 1 mole of carbon-dioxide (gas)  weighing 44g.

 

 

Question 1.4    Calculate the amount of carbon dioxide that could be produced when

                (ii) 1 mole of carbon is burnt in 16 g of dioxygen.

Answer:

When carbon is burnt in 16 g of dioxygen:

The chemical equation for this reaction is:

C_{(s)}+O_{2(g)} \rightarrow CO_{2(g)}

Here, oxygen acts as a limiting reagent as only 16g of dioxygen is available.

Hence, it will react with 0.5 mole of carbon to give 22g of carbon dioxide.

Question 1.4    Calculate the amount of carbon dioxide that could be produced when

                (iii) 2 moles of carbon are burnt in 16 g of dioxygen.

Answer:

When 2 moles of carbon is burnt in 16 g of dioxygen:

The chemical equation for this reaction is:

C_{(s)}+O_{2(g)} \rightarrow CO_{2(g)}

Here, oxygen acts as a limiting reagent as only 16g of dioxygen is available.

Thus, 16g of dioxygen will react with 0.5 mole of carbon to give 22g of carbon dioxide.

 

Question 1.5  Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol –1.

Answer:

0.375 molar aqueous solution would contain 0.375 moles of CH_{3}COONa dissolved in 1000mL of solvent.

So, we have to calculate for 500mL solution of CH_{3}COONa.

Therefore the number of moles of sodium acetate in 500mL will be:

= \frac{0.375}{1000}\times 500

=0.1875\ mole

Given Molar mass of sodium acetate: 82.0245\ g\ mol^{-1}

So, the required mass of sodium acetate = (82.0245\ g\ mol^{-1})\times (0.1875\ mole)

= 15.38\ grams

 

Question 1.6   Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL-1 and the mass percent of nitric acid in it being 69%.

Answer:

Given the mass percentage of nitric acid is 69\%.

That means 69 grams of nitric acid are present in 100 grams of nitric acid solution.

The molar mass of nitric acid HNO_{3} is 1+14+ 3\times (16) = 63g\ mol^{-1}.

So, the number of moles in 69g of Nitric acid:

  = \frac{69}{63}\ moles = 1.095\ moles

and volume of 100g of the nitric acid solution:

=\frac{100}{1.41}mL = 70.92mL = 0.07092\ L

Therefore, the concentration of Nitric acid in moles per litre is:

= \frac{1.095}{0.07092} = 15.44\ M

 

Question 1.7    How much copper can be obtained from 100 g of copper sulphate (CuSO4)?

Answer:

Given that 100g of Copper sulphate CuSO_{4};

1 mole of CuSO_{4} contains 1 mole of copper.

Molar Mass of Copper sulphate is:

= 63.5 + 32.00 + 4\times 16.00 = 159.5g

Now, 159.5g of copper sulphate contains 63.5g of copper.

So, 100g of copper sulphate will contain copper content:

 =\frac{63.5}{159.5}\times 100 = 39.81g

 

Question 1.8    Determine the molecular formula of an oxide of iron, in which the mass percent of iron and oxygen are 69.9 and 30.1, respectively. Given that the molar mass of the oxide is 159.69 g\ mol^{-1}.

Answer:

Given that the mass percentage of iron is 69.9\% and the mass percentage of oxygen is 30.1\%

The atomic mass of iron = 55.85\ u.

The atomic mass of oxygen =16.00\ u.

So, the relative moles of iron in iron oxide will be:

=\frac{mass\ of\ iron\ by\ mass}{Atomic\ mass\ of\ iron}

= \frac{69.9}{55.85} = 1.25

And relative moles of oxygen in iron oxide will be:

=\frac{mass\ of\ oxygen \ by\ mass}{Atomic\ mass\ of\ oxygen}

= \frac{30.01}{16.00} = 1.88

Hence the simplest molar ratio:

=\frac{1.25}{1.88}   or   \Rightarrow 1:1.5 = 2:3

Therefore, the empirical formula of iron oxide will be Fe_{2}O_{3}.

Now, calculating the molar mass of Fe_{2}O_{3}:

=(2\times 55.85) + (3\times16.00) = 159.7g\ mol^{-1}.

Hence it is matching with the given molar mass of the oxide.

 

Question 1.9    Calculate the atomic mass (average) of chlorine using the following data:    

                            % Natural Abundance                 Molar Mass
35Cl                                    75.77                                      34.9689
37Cl                                    24.23                                      36.9659

Answer:

To calculate the average atomic mass of chlorine:

Given the fractional natural abundance of _{}^{35}\textrm{Cl} with 34.9689\ u molar mass is 75.77\% and that of  _{}^{37}\textrm{Cl} with 36.9659\ u molar mass is 24.23\%.

Therefore we have,

Average Atomic mass of Chlorine:

= \left ( 0.7577\times34.9689\ u \right ) + \left ( 0.2423\times36.9659\ u \right )

= 26.4959\ u + 8.9568\ u = 35.4527\ u

 

Question 1.10    In three moles of ethane (C2H6), calculate the following:

                (i) Number of moles of carbon atoms.

Answer:

Given there are three moles of Ethane C_{2}H_{6};

So, 1 mole of C_{2}H_{6} contains 2 moles of carbon atoms.

Therefore, 3 moles of C_{2}H_{6} contains 6 moles of carbon atoms.

 

Question 1.10    In three moles of ethane (C2H6), calculate the following:

                (ii) Number of moles of hydrogen atoms.

Answer:

Given there are three moles of Ethane C_{2}H_{6};

So, 1 mole of C_{2}H_{6} contains 6 moles of hydrogen atoms.

Therefore, 3 moles of C_{2}H_{6} contains 18 moles of hydrogen atoms.

 

Question 1.10    In three moles of ethane (C2H6), calculate the following:

                (iii) Number of molecules of ethane.

Answer:

Given there are three moles of Ethane C_{2}H_{6};

So, 1 mole of C_{2}H_{6} contains 6.02\times 10^{23} molecules of ethane.

Therefore, 3 moles of C_{2}H_{6} contains 3\times6.02\times 10^{23} = 18.06\times10^{23} molecules of ethane.

 

Question 1.11    What is the concentration of sugar (C12H22O11) in mol L–1 if its 20 g are dissolved in enough water to make a final volume up to 2L?

Answer:

The molar mass of sugar  C_{12}H_{22}O_{11} is:

= (12\times12)+ (1\times 22)+(11\times16) = 342g\ mol^{-1}.

The number of moles of sugar in 20g of sugar will be:

= \frac{20}{342} = 0.0585\ mole

and given the volume of the solution after dissolving enough water is 2L.

 Molar\ concentration = \frac{Moles\ of\ solute}{Volume\ of\ solution\ in\ 1L}

= \frac{0.0585mol}{2L} = 0.0293mol\ L^{-1} = 0.0293\ M

 

Question 1.12    If the density of methanol is 0.793 kg L–1, what is its volume needed for making 2.5 L of its 0.25 M solution?

Answer:

Given that the density of methanol CH_{3}OH is 0.793 Kg\ L^{-1}.

So, the number of moles present in the methanol per litre will be or the Molarity of the solution will be : 

= \frac{0.793Kg\ L^{-1}}{0.032Kg\ mol^{-1}} = 24.78\ mol\ L^{-1}

Now, to make 2.5L of its 0.25M solution:

We will apply the formula: M_{1}V_{1} (for\ given\ solution) = M_{2}V_{2} (for\ solution\ to\ be\ prepared)

24.78\times V_{1} =0.25\times 2.5L

V_{1} = 0.02522L\ or\ 25.22\ mL.

Hence 25.22\ mL volume will be required for making  2.5L of methanol 0.25M solution.

 

Question 1.13    Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below:
            1 Pa = 1 Nm–2
If mass of air at sea level is 1034 g cm–2, calculate the pressure in pascal.

Answer:

The pressure is as given is force per unit area of the surface.

 Pressure = \frac{Force}{Area}

Given to calculate the pressure exerted by the air on sea water, if the mass of air at sea level is 1034\ g\ cm^{-2}.

The force with which the air is exerting on the surface is:

= \frac{1034g\times9.8ms^{-2}}{cm^2}\times\frac{1kg}{1000g}\times\frac{100cm}{1m} = 1.01332\times10^5\ N

Now,

as 1 Pascal = 1N\ m^{-2}

Therefore, 1.01332\times10^5N\timesm^{-2} = 1.01332\times10^{5}\ Pa

 

Question 1.14    What is the SI unit of mass? How is it defined?

Answer:

The SI unit of mass is Kilogram (Kg).

It is defined as the mass of platinum-iridium (Pt-Ir) cylinder that is stored in an air-tight jar at the International Bureau of Weigh and Measures in France.

 

Question 1.15    Match the following prefixes with their multiples:

                Prefixes                 Multiples

                (i) micro                      106
                (ii) deca                       109
                (iii) mega                    10–6
                (iv) giga                      10–15
                (v) femto                    10

Answer:

Matched items are given in below table:

Prefixes

Multiples

(i) micro

10^{-6}

(ii) Deca

10

(iii) Mega

10^6

(iv) Giga

10^9

(v) femto

10^{-15}

 

Question 1.16    What do you mean by significant figures?

Answer:

Significant figures are meaningful digits that are known with certainty including the last digit whose value is uncertain.

For example: if we write a result as 56.923 Kg, we say the 56.92 is certain and 3 is uncertain and the uncertainty would be \pm 1 in the last digit. Here we also include the last uncertain digit in the significant figures.

 

Question 1.17    A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).

(ii)    Determine the molality of chloroform in the water sample.

Answer:

The molarity of chloroform sample in water will be:

= \frac{number\ of\ moles\ present}{Volume\ of\ solution\ in\ L}

The molar mass of Chloroform CHCl_{3}  :

= 12+1+(3\times35.5) = 118.5g\ mol^{-1}

So we have calculated in the previous part that percentage by mass of chloroform is 1.5\times10^{-3}.

Hence in 100g sample, there will be 1.5\times10^{-3}g of chloroform.

Therefore, 1000g (1Kg) of the sample will contain choroform = 1.5\times10^{-2}g.

= \frac{1.5\times10^{-2}}{118.65\ mol} = 1.266\times 10^{-4}\ mol

Therefore, the molarity of chloroform in the water sample is 1.266\times 10^{-4}\ M.

 

Question 1.18    Express the following in the scientific notation:

                (i) 0.0048

Answer:

The scientific notation of 0.0048 will be 4.8\times 10^{-3}.

 

Question 1.18    Express the following in the scientific notation:

     (ii) 234,000

Answer:

The scientific notation of 234,000 will be 2.34\times10^5.

 

Question  1.18    Express the following in the scientific notation:

                (iii) 8008

Answer:

The scientific notation of 8008 will be 8.008\times 10^3.

 

Question 1.18    Express the following in the scientific notation:

                (iv) 500.0    

Answer:

The scientific notation of 500.0 will be 5.000\times10^2.

 

Question 1.18    Express the following in the scientific notation:

                (v) 6.0012

Answer:

The scientific notation of 6.0012 will be 6.0012\times10^0???????.

 

Question 1.19    How many significant figures are present in the following?

                (i) 0.0025

Answer:

There are 2 significant digits because all non-zero digits are in a number are significant and the zeros written to the left of the first non-zero digit in a number are non-significant.

 

Question 1.19    How many significant figures are present in the following?

                (ii) 208

Answer:

There are 3 significant digits because all non-zero digits are in a number are significant and the zeros between two non-zero digits are always significant.

 

Question 1.19    How many significant figures are present in the following?

                (iii) 5005

Answer:

There are 4 significant digits because all non-zero digits are in a number are significant and the zeros between two non-zero digits are always significant.

 

Question 1.19    How many significant figures are present in the following?

                (iv) 126,000

Answer:

There are 3 significant digits because all non-zero digits are in a number are significant and the terminal zeros are not significant if there is no decimal point.

 

Question 1.19    How many significant figures are present in the following?

                (v) 500.0

Answer:

There are 4 significant digits because the zeros written to the left of the first non-zero digit in a number are non-significant and all zeros placed to the right of a decimal point in a number are significant.

 

Question 1.19    How many significant figures are present in the following?

                (vi) 2.0034

Answer:

There are 5 significant digits because all zeros placed to the right of a decimal point in a number are significant.

 

Question 1.20    Round up the following upto three significant figures:

                (i) 34.216

Answer:

After round upto three significant figures:

Answer - 34.2

 

Question 1.20    Round up the following upto three significant figures:

          (ii) 10.4107

Answer:

After round upto three significant figures:

Answer - 10.4.

 

Question 1.20    Round up the following upto three significant figures:

     (iii) 0.04597

Answer:

After round up to three significant figures: 

Here the rightmost digit to be removed is more than 5 i.e., 7, then the preceding number is increased by one. So we get 0.0460.

Answer - 0.0460

 

Question 1.20    Round up the following upto three significant figures:

                (iv) 2808

Answer:

After round upto three significant figures: 

Here the rightmost digit to be removed is more than 5 i.e., 8, then the preceding number is increased by one. So we get 2810.

Answer - 2810

 

Question 1.21    The following data are obtained when dinitrogen and dioxygen react together to form different compounds:
                        Mass of dinitrogen                 Mass of dioxygen
                            (i) 14 g                                              16 g
                            (ii) 14 g                                             32 g
                            (iii) 28 g                                            32 g
                            (iv) 28 g                                            80 g

        (b) Fill in the blanks in the following conversions:
                    (i) 1 km = ...................... mm = ...................... pm
                    (ii) 1 mg = ...................... kg = ...................... ng
                    (iii) 1 mL = ...................... L = ...................... dm3

Answer:

(i) As we know in 1km = 1000m or  1m = 1000mm.  or  1 km = 1000\times1000 mm =10^6 mm

And  1pm = 1\times10^{-12}m   or   1km = 10^{15} pm

Therefore we have,

1 km = 10^6 mm =10^{15} pm

(ii) As we know in 1kg = 1000g or  1g = 1000mg.  or  1 kg = 1000\times1000 mg   or   1mg = \frac{1}{10^6} kg=10^{-6}kg

And  1 ng = 10^{-9}g   or   1ng = 10^{-9}\times1000mg   or   1ng = 10^{-6}mg

or   1 mg =10^6ng.

Therefore we have,

1 mg = 10^{-6} kg =10^{6} ng.

(iii) As we know in 1L= 1000mL or  1mL =\frac{1}{1000}L = 10^{-3}L 

or  1dm = 0.1 m \Rightarrow 1dm = 10cm   or   1cm = 0.1dm

And 1mL = 1cm^3  or we can write it as: 1mL = \frac{1dm}{10cm}\times\frac{1dm}{10cm}\times\frac{1dm}{10cm}

And  1mL = 10^{-3}dm^3  

Therefore we have,

1mL = 10^{-3}L =10^{-3}dm^3

 

Question 1.22    If the speed of light is 3.0 \times 10^8 m s^{-1} , calculate the distance covered by light in 2.00 ns.

Answer:

Given the speed of light to be 3.0 \times 10^8 m s^{-1}, so the distance covered by light in 2.00 ns. will be:

Distance = Speed\times Time

Distance = (3.0\times10^{8}ms^{-1})\times(2.00\times10^{-9}sec) = 6.00\times10^{-1}m = 0.600m

Therefore, the light will travel 0.600 metres in 2 nano seconds.

 

Question 1.23    In a reaction
            A + B_2 \rightarrow AB_2
Identify the limiting reagent, if any, in the following reaction mixtures.

        (i) 300 atoms of A + 200 molecules of B

Answer:

Here according to the reaction 1 atom of A reacts with 1 molecule of B.

Therefore, 200 molecules of atoms of B will react with 200 atoms of A, thereby left with 100 atoms of A unreacted.

Hence, B is the limiting reagent in this reaction.

 

Question 1.23    In a reaction
            A + B_2 \rightarrow AB_2
Identify the limiting reagent, if any, in the following reaction mixtures.

            (ii) 2 mol A + 3 mol B

Answer:

Here according to the reaction 1 atom of A reacts with 1 molecule of B.

Therefore, 2 mol of  A atoms will react with only 2 mol of B molecules, thereby left with 1 mole of B unreacted.

Hence, A is the limiting reagent in this reaction.

 

Question 1.23    In a reaction
            A + B_2 \rightarrow AB_2
Identify the limiting reagent, if any, in the following reaction mixtures.

            (iii) 100 atoms of A + 100 molecules of B

Answer:

Here according to the reaction 1 atom of A reacts with 1 molecule of B.

Therefore, all 100 atoms of A will react with 100 molecules of B, so the reaction is stoichiometric and there is no limiting reagent.

Hence, there is no limiting reagent in this reaction.

 

Question 1.23    In a reaction
            A + B_2 \rightarrow AB_2
Identify the limiting reagent, if any, in the following reaction mixtures.

            (iv) 5 mol A + 2.5 mol B

Answer:

Here according to the reaction 1 atom of A reacts with 1 molecule of B.

Therefore, 2.5 moles of B molecules will react with only 2.5 moles of A atoms, thereby left with 2.5 moles of A unreacted.

Hence, B limiting reagent in this reaction.

 

Question 1.23    In a reaction
            A + B_2 \rightarrow AB_2
Identify the limiting reagent, if any, in the following reaction mixtures.

            (v) 2.5 mol A + 5 mol B

Answer:

Here according to the reaction 1 atom of A reacts with 1 molecule of B.

Therefore, 2.5 moles of A atoms will react with only 2.5 moles of B molecules, thereby left with 2.5 moles of B unreacted.

Hence, A limiting reagent in this reaction.

 

Question 1.24    Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
        N_2\ (g) + H_2\ (g) \rightarrow 2NH_3\ (g)

      (i) Calculate the mass of ammonia produced if 2.00 × 10 3 g dinitrogen reacts with 1.00 ×10 3 g of dihydrogen.

Answer:

Given that if 2.00\times10^3g of dinitrogen reacts with 1.00\times10^3g of dihydrogen.

From the reaction we have:

1 mole of dinitrogen weighing 28g reacts with 3 moles of dihydrogen weighing 6g to give 2 moles of ammonia weighing 34g.

Therefore, 2.00\times10^3g of N_{2} will react with H_{2}  = \frac{6}{28}\times200g = 428.6g.

Thus, here N_{2} is the limiting reagent while H_{2} is in excess.

So, 28g of N_{2} produces 34g of NH_{3}.

Therefore, 2.00\times10^3g of N_{2} will produce = \frac{34}{28}\times 2000g = 2428.57g of NH_{3}.

 

Question 1.24    Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
        N_2\ (g) + H_2\ (g) \rightarrow 2NH_3\ (g)

        (ii)    Will any of the two reactants remain unreacted?

Answer:

As from the previous part we have:

N_{2} is the limiting reagent and H_{2} is the excess reagent.

Hence, H_{2} will remain unreacted.

 

Question 1.25    How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?

Answer:

Calculating the molar mass of Na_{2}CO_{3}:

= (2\times23)+12.00+(3\times16) = 106g\ mol^{-1}

Therefore,

0.50\ mol of Na_{2}CO_{3} means:

0.50\times106g = 53g

whereas, 

0.50 M Na_{2}CO_{3} means:

0.50\ mol of Na_{2}CO_{3} or 53g of Na_{2}CO_{3} are present in 1litre of the solution.

 

Question 1.26    If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?

Answer:

For the given situation we have the reaction:

2H_{2}(g)+O_{2}(g)\rightarrow2H_{2}O(g)

Here, 2 volumes of dihydrogen gas react with 1 volume of dioxygen to produce 2 volumes of water vapour.

So, if 10 volumes of dihydrogen gas react with 5 volume of dioxygen then it will produce (2\times5 = 10  volumes of water vapour.

 

Question 1.27    Convert the following into basic units:

                (i)    28.7 pm

Answer:

To convert 28.7 pm into the basic units:

As 1pm = 10^{-12}m.

\therefore 28.7\ pm = 28.7\times10^{-12}m = 2.87\times10^{-11}m

 

Question 1.27    Convert the following into basic units:

                (ii) 15.15 pm

Answer:

To convert 15.15 pm into the basic units:

As 1pm = 10^{-12}m.

\therefore 15.15\ pm = 15.15\times10^{-12}m = 1.515\times10^{-11}m

 

Question 1.27    Convert the following into basic units:

                (iii) 25365 mg

Answer:

To convert 25365 mg into basic unit:

As 1mg = 10^{-3}g.

\therefore 25365\ mg = 2.5365\times10^4\times10^{-3}g

Now, as 

1 g= 10^{-3} kg

2.5365\times10g = 2.5365\times10\times10^{-3} kg

\therefore 25365\ mg = 2.5365\times 10^{-2}kg

 

Question 1.28    Which one of the following will have the largest number of atoms?
                (i) 1 g Au (s)
                (ii) 1 g Na (s)
                (iii) 1 g Li (s)
                (iv) 1 g of Cl2 (g)

Answer:

Calculating and then comparing for each:

(i) 1 g of Au will contain:

= \frac{1}{197}\ mol= \frac{1}{197}\times6.022\times10^{23}\ atoms.

(ii) 1 g of Na will contain:

= \frac{1}{23}\ mol= \frac{1}{23}\times6.022\times10^{23}\ atoms. 

(iii) 1 g of Li will contain:

= \frac{1}{7}\ mol= \frac{1}{7}\times6.022\times10^{23}\ atoms.

(iv) 1 g of Cl2 will contain:

= \frac{1}{71}\ mol= \frac{1}{71}\times6.022\times10^{23}\ atoms.

Clearly, we can compare and say that the number of atoms in 1g of Li has the largest.

 

Question 1.29    Calculate the molarity of a solution of ethanol in water, in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).

Answer:

Given that the mole fraction of ethanol in water is 0.040.

Mole\ fraction\ of\ ethanol\ = \frac{Number\ of\ moles\ of\ ethanol}{Total\ number\ of\ moles\ of\ solution}

\Rightarrow X_{ethanol} = \frac{n_{ethanol}}{n_{water}+n_{ethanol}} = 0.040

To find the molarity we must have to find the number of moles of ethanol present in 1litre of solution.

Assuming the density of water to be 1 kg\ m^3.

Therefore, water is approximately equal to 1Litre.

The number of moles in 1L of water:

= \frac{1000g}{18g}\ mol^{-1} = 55.55\ moles

So, substituting in place of n_{water} = 55.55 in above equation we get,

\Rightarrow X_{ethanol} = \frac{n_{ethanol}}{55.55+n_{ethanol}} = 0.040

\Rightarrow 0.96\times n_{ethanol} = 55.55\times0.040

n_{ethanol} = 2.31\ mol

So, 2.31\ moles are present in 1L of solution.

Hence, the molarity of the solution is 2.31\ M

 

Question 1.30    What will be the mass of one 12C atom in g?

Answer:

As we know the mass of 1 mole of _{}^{12}C\textrm{} atoms or 6.022\times10^{23} number of atoms is 12grams. Hence, the mass of one _{}^{12}C\textrm{} atom will be:\Rightarrow \frac{12}{6.022\times10^{23}g} = 1.9927\times10^{-23}g

 

Question 1.31    How many significant figures should be present in the answer of the following calculation?

                (i)    \frac{0.02856\times 298.12\times 0.112}{0.5785}

Answer:

To find the number of significant figures that would be present in the answer, we will be finding the least precise term, having the least significant figures.

Here, the least precise term is 0.112 having only 3 significant digits.

Therefore, there will be 3 significant figures in the calculated answer.

 

Question 1.31    How many significant figures should be present in the answer of the following calculation?

        (ii)    5\times 5.364

Answer:

Here, 5.364 is having 4 significant digits.

Therefore, after multiplying by 5 the answer would also have the same 4 significant figures.

 

Question 1.31    How many significant figures should be present in the answer of the following calculation?

      (iii)    0.0125 + 0.7864 + 0.0215

Answer:

Here, the least number of decimal places in each term is four.

Therefore, the calculation would also have the same 4 significant figures.

 

Question 1.32    Use the data given in the following table to calculate the molar mass of naturally occuring argon isotopes:

Isotope

Isotopic molar mass

Abundance

36Ar

35.96755 g mol-1

0.337%

38Ar

37.96272 g mol-1

0.063%

40Ar

39.9624 g mol-1

99.600%

 

Answer:

For different isotopes of argon, we have given their naturally occurring abundances.

So, to calculate the molar mass:

Multiply the isotopic molar mass with their abundance to get the molar mass, and then add all of them to get,

\therefore Molar\ Mass\ of Argon = \sum m_{i}A_{i}

= (35.96755\times0.337)+ (37.96272\times0.063)+(39.9624\times99.600)

= 39.948g\ mol^{-1}

 

Question 1.33    Calculate the number of atoms

     (i) 52 moles of Ar

Answer:

As 1 mole of air contains 6.022\times10^{23} atoms.

Therefore, 52 moles of Ar will contain 52\times6.022\times10^{23} atoms.

\Rightarrow 3.131\times10^{25}\ atoms.

 

Question 1.33    Calculate the number of atoms

       (ii) 52 u of He.

Answer:

As 1 atom of He weights 4 u.

Hence the number of atoms of He present in 52 u will be:

=\frac{52\ u}{4\ u} = 13\ atoms

 

Question 1.33    Calculate the number of atoms

     (iii) 52 g of He.

Answer:

As 1 mole of He weights 4g and contains 6.022\times10^{23} atoms.

Therefore, 52g will contain \frac{52}{4}moles = 13\ moles .

Hence, the number of atoms will be 13\times6.022\times10^{23}\ atoms.

\Rightarrow 7.8286\times10^{24}\ atoms.

 

Question 1.34    A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate

  (i) empirical formula

Answer:

The amount of carbon in 3.38 g of CO_{2} :

= \frac{12}{44}\times3.38g = 0.9218g

The amount of hydrogen in 0.690g of H_{2}O:

= \frac{2}{18}\times0.690g = 0.0767g

The compound contains only C and H, 

Therefore, the total mass of the compound will be:

= 0.9218+0.0767 = 0.9985g

Now, the percentage of Carbon in the compound:

= (\frac{0.9218}{0.9985})\times100 = 92.32

and the percentage of Hydrogen in the compound:

= (\frac{0.0767}{0.9985})\times100 = 7.68

Now, the empirical formula,

Moles of carbon in the compound:

= \frac{92.32}{12} = 7.69

Moles of hydrogen in the compound:

= \frac{7.68}{1} = 7.68

So, the simplest molar ratio will be = 7.69:7.68 = 1:1

Therefore, the empirical formula is CH.

 

Question 1.35    Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction,

    CaCO_3 (s) + 2 HCl (aq) \rightarrow CaCl_2 (aq) + CO_2 (g) + H_2 O(l)

What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?

Answer:

0.75M HCl contains 0.75\ mole in 1000mL of solution.

Or,  0.75\times36.5g = 24.375g of HCl in 1000mL solution.

Therefore,

Mass of HCl in 25mL of 0.75M\ HCl:

= \frac{24.375}{1000}\times25 g = 0.6844g

so, from the given chemical equation,

CaCO_3 (s) + 2 HCl (aq) \rightarrow CaCl_2 (aq) + CO_2 (g) + H_2 O(l)

1 mole of CaCO_{3}(s) i.e., 100g reacts with 2 moles of HCl(aq) i.e., 73g.

Therefore, 0.6844g HCl reacts completely with CaCO_{3} to give:

=\frac{100}{73}\times 0.6844g = 0.938 g

 

Question 1.36    Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction

         4 HCl (aq) + MnO_2 (s) \rightarrow 2H_2 O (l) + MnCl_2 (aq) + Cl_2 (g)

How many grams of HCl react with 5.0 g of manganese dioxide?

Answer:

Molar mass of MnO_{2} is 55+32 g = 87g.

Here from the reaction 1 mole of MnO_{2} reacts with 4 moles of HCl,

i.e., 4\times36.5g =146g of HCl.

Therefore, 5.0 g of MnO_{2} will react with HCl:

= \frac{146}{87}\times5.0g = 8.40g.

 

NCERT solutions for class 11 chemistry

Chapter 1

NCERT solutions for class 11 chemistry chapter 1 Some Basic Concepts of Chemistry

Chapter-2

CBSE NCERT solutions for class 11 chemistry chapter 2 Structure of Atom

Chapter-3

Solutions of NCERT class 11 chemistry chapter 3 Classification of Elements and Periodicity in Properties

Chapter-4

NCERT solutions for class 11 chemistry chapter 4 Chemical Bonding and Molecular Structure

Chapter-5

CBSE NCERT solutions for class 11 chemistry chapter 5 States of Matter

Chapter-6

Solutions of NCERT class 11 chemistry chapter 6 Thermodynamics

Chapter-7

NCERT solutions for class 11 chemistry chapter 7 Equilibrium

Chapter-8

CBSE NCERT solutions for class 11 chemistry chapter 8 Redox Reaction

Chapter-9

Solutions of NCERT class 11 chemistry chapter 9 Hydrogen

Chapter-10

NCERT solutions for class 11 chemistry chapter 10 The S-Block Elements

Chapter-11

CBSE NCERT solutions for class 11 chemistry chapter 11 The P-Block Elements

Chapter-12

Solutions of NCERT class 11 chemistry chapter 12 Organic chemistry- some basic principles and techniques

Chapter-13

NCERT solutions for class 11 chemistry chapter 13 Hydrocarbons

Chapter-14

CBSE NCERT solutions for class 11 chemistry chapter 14 Environmental Chemistry

NCERT solutions for class 11 Subject wise

NCERT solutions for class 11 biology

Solutions of NCERT class 11 maths

CBSE NCERT solutions for class 11 chemistry

NCERT solutions for class 11 physics

Benefits of NCERT solutions for class 11 chemistry chapter 1 Some Basic Concepts of Chemistry

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