# 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

The molar mass of Water $(H_{2}O)$ is:

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

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

$= 18.02\ u$

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

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

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

$= 44.011\ u$

The molar mass of Methane $CH_{4}$ is:

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

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

$= 16.043\ u$

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):

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

$= 45.049 \%$

$\approx 45.0\%$

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 $Fe_{2}O_{3}$.

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

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.

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

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.

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

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.

$0.375$ 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 $= (82.0245\ g\ mol^{-1})\times (0.1875\ mole)$

$= 15.38\ grams$

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:

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

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:

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

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.

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

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

(i) Number of moles of carbon atoms.

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

(ii) Number of moles of hydrogen atoms.

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

(iii) Number of molecules of ethane.

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.

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.

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

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

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.

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

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

Prefixes                 Multiples

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

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

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.

(i) Express this in per cent by mass.

1ppm means 1 parts in million $(10^6)$ parts.

So, in percentage by mass:

We have,

$Percentage\ by\ mass = \frac{15}{10^6}\times 100 = 15\times10^{-4}$

$= 1.5\times10^{-3}\%$

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

(i) 0.0048

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

Question

(ii) 234,000

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

Question

(iii) 8008

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

Question

(iv) 500.0

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

Question

(v) 6.0012

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

Question

(i) 0.0025

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

(ii) 208

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

(iii) 5005

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

(iv) 126,000

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

(v) 500.0

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

(vi) 2.0034

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

Question

(i) 34.216

After round upto three significant figures:

Answer - $34.2$

Question

(ii) 10.4107

After round upto three significant figures:

Answer - $10.4$.

Question

(iii) 0.04597

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

(iv) 2808

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 - $2810$

Here if we fix the mass of dinitrogen at 14g, then the masses of dioxygen that will combine with the fixed mass of dinitrogen are 16g, 32g, 32g, and 80g.

The masses of dioxygen bear a whole number ratio of 1:2:2:5.

Hence, the given experimental data obeys the Law of Multiple Proportionals.

The law given by Dalton states that "if two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers".

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

(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,

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

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.

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

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.

(ii) 2 mol A + 3 mol B

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.

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

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.

(iv) 5 mol A + 2.5 mol B

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.

(v) 2.5 mol A + 5 mol B

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.

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

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 $NH_{3}$.

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

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.

(iii) If yes, which one and what would be its mass?

Yes$H_{2}$ will remain unreacted.

And the mass of dihydrogen left unreacted will be $=1000g-428.6g = 571.4g$

Question

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.

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

(i)    28.7 pm

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

(ii) 15.15 pm

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

(iii) 25365 mg

To convert 25365 mg into basic unit:

As $1mg = 10^{-3}g$.

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

Now, as

$1 g= 10^{-3} kg$

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

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

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.

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

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:$\Rightarrow \frac{12}{6.022\times10^{23}g} = 1.9927\times10^{-23}g$

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

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.

(ii)    $5\times 5.364$

Here, $5.364$ is having 4 significant digits.

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

(iii)    0.0125 + 0.7864 + 0.0215

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

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

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%

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

(i) 52 moles of Ar

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

(ii) 52 u of He.

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

(iii) 52 g of He.

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 $13\times6.022\times10^{23}\ atoms$.

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

(i) empirical formula

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 $CH$.

(ii) molar mass of the gas

Wight of 10 Litres of the gas at S.T.P. is $11.6g$

Weight of 22.4 Litres of gas at S.T.P. will be:

$= \frac{11.6g}{10.0L}\times22.4L$

$= 25.984\ g\ mol^{-1}$

$\approx 26\ g\ mol^{-1}$

(iii) molecular formula.

As we know the empirical formula from the previous part is $CH$.

The mass of empirical formula $CH$ $= 12+1 =13$.

Therefore,

$n = \frac{Molecular\ Mass}{Empirical\ Formula}$

$\Rightarrow \frac{26}{13} =2$

Therefore, Molecular formula is $C_{2}H_{2}$.

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

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

$0.75M$ $HCl$ 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:

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

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

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

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

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