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

Ans : (i) \( \mathrm{H}_{2} \mathrm{O} \):
The molecular mass of water, \( \mathrm{H}_{2} \mathrm{O} \)
\( =(2 \times \text { Atomic mass of hydrogen) }+(1 \times \text { Atomic mass of oxygen }) \)
\( =[2(1.0084)+1(16.00 \mathrm{u})] \)
\( =2.016 \mathrm{u}+16.00 \mathrm{u} \)
\( \begin{aligned} &=18.016 \\ &=18.02 \mathrm{u} \end{aligned} \)

(ii) \( \mathrm{CO}_{2} \)
The molecular mass of carbon dioxide, \( \mathrm{CO}_{2} \)
\( \begin{aligned} &=(1 \times \text { Atomic mass of carbon })+(2 \times \text { Atomic mass of oxygen }) \\ &=[1(12.011 \mathrm{u})+2(16.00 \mathrm{u})] \\ &=12.011 \mathrm{u}+32.00 \mathrm{u} \\ &=44.01 \mathrm{u} \end{aligned} \)

(iii) \( \mathrm{CH}_{4 :} \)
The molecular mass of methane, \( \mathrm{CH}_{4} \)
\( \begin{array}{l}{=(1 \times \text { Atomic mass of carbon })+(4 \times \text { Atomic mass of hydrogen })} \\ {=[1(12.011 \mathrm{u})+4(1.008 \mathrm{u})]} \\ {=12.011 \mathrm{u}+4.032 \mathrm{u}} \\ {=16.043 \mathrm{u}}\end{array} \) 

Ans : The molecular formula of sodium sulphate is \( \mathrm{Na}_{2} \mathrm{SO}_{4} \)
Molar mass of \( \mathrm{Na}_{2} \mathrm{SO}_{4} \) \( =[(2 \times 23.0)+(32.066)+4(16.00)] \)
\( =142.066 \mathrm{g} \)
\( \text{Mass percent of an element} \) \( =\frac{\text { Mass of that element in the compound }}{\text { Molar mass of the compound }} \times 100 \)
\( \therefore \) Mass percent of sodium:
\( =\frac{46.0 \mathrm{g}}{142.066 \mathrm{g}} \times 100 \)
\( \begin{array}{l}{=32.379} \\ {=32.4 \%}\end{array} \)
Mass percent of sulphur:
\( =\frac{32.066 \mathrm{g}}{142.066 \mathrm{g}} \times 100 \)
\( \begin{aligned} &=22.57 \\ &=22.6 \% \end{aligned} \)
Mass percent of oxygen:
\( =\frac{64.0 \mathrm{g}}{142.066 \mathrm{g}} \times 100 \)
\( \begin{array}{l}{=45.049} \\ {=45.05 \%}\end{array} \) 

Ans : \( \% \text { of iron by mass }=69.9 \%[\text { Given }] \)
\( \% \text { of oxygen by mass }=30.1 \%[\text { Given }] \)
\( \text{Relative moles of iron in iron oxide:} \)
\( =\frac{\% \text { of iron by mass }}{\text { Atomic mass of iron }} \)
\( =\frac{69.9}{55.85} \)
\( \begin{array}{l}{=1.25} \\ {\text { Relative moles of oxygen in iron oxide: }}\end{array} \)
\( =\frac{\% \text { of oxygen by mass }}{\text { Atomic mass of oxygen }} \)
\( =\frac{30.1}{16.00} \)
\( =1.88 \)
Simplest molar ratio of iron to oxygen:
\( \begin{array}{l}{=1.25 : 1.88} \\ {=1 : 1.5} \\ {=2 : 3}\end{array} \)
\( \therefore \text { The empirical formula of the iron oxide is } \mathrm{Fe}_{2} \mathrm{O}_{3} \) 

Ans : The balanced reaction of combustion of carbon can be written as:
(i) As per the balanced equation, 1 mole of carbon burns in 1 mole of dioxygen (air) to Produced 1 mole of carbon dioxide.
(ii) According to the question, only 16 g of dioxygen is available. Hence, it will react with 0.5 mole of carbon to give 22 g of carbon dioxide. Hence, it is a limiting reactant.
(iii) According to the question, only 16 g of dioxygen is available. It is a limiting reactant. Thus, 16 g of dioxygen can combine with only 0.5 mole of carbon to give 22 g of carbon dioxide. 

Ans : \( 0.375 \mathrm{M} \) aqueous solution of sodium acetate.
\( \equiv 1000 \mathrm{mL} \text { of solution containing } 0.375 \) moles of sodium acetate.
\( \therefore \) Number of moles of sodium acetate in 500 mL.
\( =\frac{0.375}{1000} \times 500 \)
\( =0.1875 \) mole
Molar mass of sodium acetate = \( 82.0245 \mathrm{g} \mathrm{mole}^{-1}(\text { Given }) \)
\( \therefore \text { Required mass of sodium acetate }=\left(82.0245 \mathrm{g} \mathrm{mol}^{-1}\right)(0.1875 \text { mole) } \)
\( =15.38 \mathrm{g} \) 

Ans : Mass percent of nitric acid in the sample \( =69 \%[\text { Given }] \)
Thus, \( 100 \mathrm{g} \text { of nitric acid contains } 69 \mathrm{g} \) of nitric acid by mass. Molar mass of nitric acid \( \left(\mathrm{HNO}_{3}\right) \)
\( =\{1+14+3(16)\} \mathrm{g} \mathrm{mol}^{-1} \)
\( \begin{array}{l}{=1+14+48} \\ {=63 \mathrm{g} \mathrm{mol}^{-1}}\end{array} \)
\( \therefore \text { Number of moles in } 69 \text { g of HNO }_{3} \)
\( =\frac{69 \mathrm{g}}{63 \mathrm{gmol}^{-1}} \)
\( =1.095 \mathrm{mol} \)
Volume of 100g nitric acid solution
\( =\frac{\text { Mass of solution }}{\text { density of solution }} \)
\( =\frac{100 \mathrm{g}}{1.41 \mathrm{g} \mathrm{mL}^{-1}} \)
\( =\frac{100 \mathrm{g}}{1.4 \lg \mathrm{mL}^{-1}} \)
\( =70.92 \mathrm{mL} \equiv 70.92 \times 10^{-3} \mathrm{L} \)
Concentration of nitric acid
\( =\frac{1.095 \mathrm{mole}}{70.92 \times 10^{-3} \mathrm{L}} \)
\( =15.44 \mathrm{mol} / \mathrm{L} \)
\( \therefore \text { Concentration of nitric acid }=15.44 \mathrm{mol} / \mathrm{L} \) 

Ans : 1 mole of \( \mathrm{CuSO}_{4} \) contains 1 mole of copper.
Molar mass of \( \operatorname{cus} O_{4}=(63.5)+(32.00)+4(16.00) \)
\( \begin{array}{l}{=63.5+32.00+64.00} \\ {=159.5 \mathrm{g}}\end{array} \)
\( 159.5 \mathrm{g} \) of \( \mathrm{CuSO}_{4} \) contains 53.5 g of copper.
\( \Rightarrow 100 \mathrm{g} \) of \( \mathrm{cusO}_{4} \) will contain \( \frac{63.5 \times 100 \mathrm{g}}{159.5} \) of copper.
\( \therefore \) AMount of copper that can be obtained from 100 g \( \mathrm{CuSO}_{4} \) \( =\frac{63.5 \times 100}{159.5} \)
\( =39.81 \mathrm{g} \) 

Ans : Mass percent of iron (Fe) = \( =69.9 \% \) (Given)
Mass percent of oxygen (O) = \( 30.1 \%(\text { Given }) \)
Number of moles of iron present in the oxide = \( \frac{69.90}{55.85} \)
\( =1.25 \)
Number of moles of oxygen present in the oxide \( =\frac{30.1}{16.0} \)
\( =1.88 \)
\( Ratio of iron to oxygen in the oxide, \)
\( =1.25 : 1.88 \)
\( =\frac{1.25}{1.25} : \frac{1.88}{1.25} \)
\( \begin{array}{l}{=1 : 1.5} \\ {=2 : 3}\end{array} \)
\( \therefore \text { The empirical formula of the oxide is } \mathrm{Fe}_{2} \mathrm{O}_{3} \)
Empirical formula mass of \( \mathrm{Fe}_{2} \mathrm{O}_{3}=[2(55.85)+3(16.00)] \mathrm{g} \)
Molar mass of \( \mathrm{Fe}_{2} \mathrm{O}_{3}=159.69 \mathrm{g} \)
\( \therefore n=\frac{\text { Molar mass }}{\text { Empirical formula mass }} \) \( =\frac{159.69 \mathrm{g}}{159.7 \mathrm{g}} \)
\( \begin{array}{l}{=0.999} \\ {=1(\text { approx })}\end{array} \)
\( \begin{array}{l}{\text { Molecular formula of a compound is obtained by multiplying the empirical formula with } n .} \\ {\text { Thus, the empirical formula of the given oxide is } \mathrm{Fe}_{2} \mathrm{O}_{3} \text { and } n \text { is } 1 \text { . }} \\ {\text { Hence, the molecular formula of the oxide is } \mathrm{Fe}_{2} \mathrm{O}_{3 .}}\end{array} \) 

Ans : The average atomic mass of chlorine.

\( \begin{array}{l}{=26.4959+8.9568} \\ {=35.4527 \mathrm{u}}\end{array} \)
\( \therefore \text { The average atomic mass of chlorine }=35.4527 \mathrm{u} \) 

Ans : \( \begin{array}{l}{\text { (i) 1 mole of } \mathrm{C}_{2} \mathrm{H}_{6} \text { contains } 2 \text { moles of carbon atoms. }} \\ {\therefore \text { Number of moles of carbon atoms in } 3 \text { moles of } \mathrm{C}_{2} \mathrm{H}_{6}} \\ {=2 \times 3=6}\end{array} \)
\( \begin{array}{l}{\text { (ii) } 1 \text { mole of } \mathrm{C}_{2} \mathrm{H}_{6} \text { contains } 6 \text { moles of hydrogen atoms. }} \\ {\therefore \text { Number of moles of carbon atoms in } 3 \text { moles of } \mathrm{C}_{2} \mathrm{H}_{6}} \\ {=3 \times 6=18}\end{array} \)
\( \begin{array}{l}{\text { (iii) } 1 \text { mole of } \mathrm{C}_{2} \mathrm{H}_{6} \text { contains } 6.023 \times 10^{23} \text { molecules of ethane. }} \\ {\therefore \text { Number of molecules in } 3 \text { moles of } \mathrm{C}_{2} \mathrm{H}_{6}} \\ {=3 \times 6.023 \times 10^{23}=18.069 \times 10^{23}}\end{array} \) 

Ans : Molarity (M) of a solution is given by,
\( =\frac{\text { Number of moles of solute }}{\text { Volume of solution in Litres }} \)
\( =\frac{\text { Mass of sugar/molar mass of sugar }}{2 L} \)
\( =\frac{20 \mathrm{g} /[(12 \times 12)+(1 \times 22)+(11 \times 16)] \mathrm{g}}{2 \mathrm{L}} \)
\( =\frac{20 \mathrm{g} / 342 \mathrm{g}}{2 \mathrm{L}} \)
\( =\frac{0.0585 \mathrm{mol}}{2 \mathrm{L}} \)
\( =0.02925 \mathrm{mol} \mathrm{L}^{-1} \)
\( \therefore \text { Molar concentration of sugar }=0.02925 \mathrm{mol} \mathrm{L}^{-1} \) 

Ans : Molar mass of methanol \( \left(\mathrm{CH}_{3} \mathrm{OH}\right)=(1 \times 12)+(4 \times 1)+(1 \times 16) \)
\( =32 \mathrm{g} \mathrm{mol}^{-1} \)
\( =0.032 \mathrm{kg} \mathrm{mol}^{-1} \)
Molarity of methanol solution \( =\frac{0.793 \mathrm{kg} \mathrm{L}^{-1}}{0.032 \mathrm{kg} \mathrm{mol}^{-1}} \)
\( =24.78 \mathrm{mol} \mathrm{L}^{-1} \)
Since density is mass per unit volume
Applying,
\( M_{1} V_{1}=M_{2} V_{2} \)
Given  solution) (Solution to be prepared
\( \left(24.78 \mathrm{mol} \mathrm{L}^{-1}\right) \mathrm{V}_{1}=(2.5 \mathrm{L})\left(0.25 \mathrm{mol} \mathrm{L}^{-1}\right) \)
\( \begin{array}{l}{\mathrm{V}_{1}=0.0252 \mathrm{L}} \\ {\mathrm{V}_{1}=25.22 \mathrm{mL}}\end{array} \) 

Ans : Pressure is defined as force acting per unit area of the surface.
\( P=\frac{F}{A} \)
\( =\frac{1034 \mathrm{g} \times 9.8 \mathrm{ms}^{-2}}{\mathrm{cm}^{2}} \times \frac{1 \mathrm{kg}}{1000 \mathrm{g}} \times \frac{(100)^{2} \mathrm{cm}^{2}}{1 \mathrm{m}^{2}} \)
\( =1.01332 \times 10^{5} \mathrm{kg} \mathrm{m}^{-1} \mathrm{s}^{-2} \)
We know,
\( \begin{array}{l}{1 \mathrm{N}=1 \mathrm{kg} \mathrm{ms}^{-2}} \\ {\text { Then, }}\end{array} \)
\( \begin{array}{l}{1 \mathrm{Pa}=1 \mathrm{Nm}^{-2}=1 \mathrm{kg} \mathrm{m}^{-2} \mathrm{s}^{-2}} \\ {1 \mathrm{Pa}=1 \mathrm{kg} \mathrm{m}^{-1} \mathrm{s}^{-2}}\end{array} \)
\( \therefore \text { Pressure }=1.01332 \times 10^{5} \mathrm{Pa} \) 

Ans : The SI unit of mass is kilogram (kg). 1 kilogram is defined as the mass equal to the mass of the international prototype of kilogram. 

Ans :  

Ans : Significant figures are those meaningful digits that are known with certainty. They indicate uncertainty in an experiment or calculated value. For example, if 15.6 mL is the result of an experiment, then 15 is certain while 6 is uncertain, and the total number of significant figures are 3. Hence, significant figures are defined as the total number of digits in a number including the last digit that represents the uncertainty of the result. 

Ans : (i) \( 1 \text { ppm is equivalent to } 1 \text { part out of } 1 \text { million }\left(10^{6}\right) \) parts.
\( \therefore \text { Mass percent of } 15 \mathrm{ppm} \) chloroform in water.
\( =\frac{15}{10^{6}} \times 100 \)
\( \approx 1.5 \times 10^{-3} \% \)

(ii) \( 100 \mathrm{g} \text { of the sample contains } 1.5 \times 10^{-3} \mathrm{g} \text { of } \mathrm{CHCl}_{3 .} \)
\( \Rightarrow 1000 \mathrm{g} \text { of the sample contains } 1.5 \times 10^{-2} \mathrm{g} \text { of } \mathrm{CHCl}_{3} \)
\( \therefore \)Molality of chloroform in water.
\( =\frac{1.5 \times 10^{-2} \mathrm{g}}{\text { Molar mass of } \mathrm{CHCl}_{3}} \)
Molar mass of \( \mathrm{CHCl}_{3}=12.00+1.00+3(35.5) \)
\( =119.5 \mathrm{g} \mathrm{mol}^{-1} \)
\( \therefore \text { Molality of chloroform in water }=0.0125 \times 10^{-2} \mathrm{m} \)
\( =1.25 \times 10^{-4} \mathrm{m} \) 

Ans : (i) \( 0.0048=4.8 \times 10^{-3} \)
(ii) \( 234,000=2.34 \times 10^{5} \)
(iii) \( 8008=8.008 \times 10^{3} \)
(iv) \( 500.0=5.000 \times 10^{2} \)
(v) \( 6.0012=6.0012 \) 

Ans : (i) There are 2 significant figures.
(ii) There are 3 significant figures.
(iii) There are 4 significant figures.
(iv) There are 3 significant figures.
(v) There are 4 significant figures.
(vi) There are 5 significant figures. 

Ans : (i) 34.2
(ii) 10.4
(iii) 0.0460
(iv) 2810 

Ans : (a) If we fix the mass of dinitrogen at 28 g, then the masses of dioxygen that will combine with the fixed mass of dinitrogen are 32 g, 64 g, 32 g, and 80 g. The masses of dioxygen bear a whole number ratio of 1:2:2:5. Hence, the given experimental data obeys the law of multiple proportions. The law states that if two elements combine to form more than one compound, then the masses of one element that combines with the fixed mass of another element are in the ratio of small whole numbers.
(b) (i) \( 1 \mathrm{km}=1 \mathrm{km} \times \frac{1000 \mathrm{m}}{1 \mathrm{km}} \times \frac{100 \mathrm{cm}}{1 \mathrm{m}} \times \frac{10 \mathrm{mm}}{1 \mathrm{cm}} \)
\( \therefore 1 \mathrm{km}=10^{6} \mathrm{mm} \)
\( 1 \mathrm{km}=1 \mathrm{km} \times \frac{1000 \mathrm{m}}{1 \mathrm{km}} \times \frac{1 \mathrm{pm}}{10^{-12} \mathrm{m}} \)
\( \therefore 1 \mathrm{km}=10^{15} \mathrm{pm} \)
Hence , \( 1 \mathrm{km}=10^{6} \mathrm{mm}=10^{15} \mathrm{pm} \)

(ii) \( 1 \mathrm{mg}=1 \mathrm{mg} \times \frac{1 \mathrm{g}}{1000 \mathrm{mg}} \times \frac{1 \mathrm{kg}}{1000 \mathrm{g}} \)
\( \Rightarrow 1 \mathrm{mg}=10^{-6} \mathrm{kg} \)
\( 1 \mathrm{mg}=1 \mathrm{mg} \times \frac{1 \mathrm{g}}{1000 \mathrm{mg}} \times \frac{1 \mathrm{ng}}{10^{-9} \mathrm{g}} \)
\( \Rightarrow 1 \mathrm{mg}=10^{6} \mathrm{ng} \)
\( \therefore 1 \mathrm{mg}=10^{-6} \mathrm{kg}=10^{6} \mathrm{ng} \)

(iii) \( 1 \mathrm{mL}=1 \mathrm{mL} \times \frac{1 \mathrm{L}}{1000 \mathrm{mL}} \)
\( \Rightarrow 1 \mathrm{mL}=10^{-3} \mathrm{L} \)
\( 1 \mathrm{mL}=1 \mathrm{cm}^{3}=1 \mathrm{cm}^{3} \frac{1 \mathrm{dm} \times 1 \mathrm{dm} \times 1 \mathrm{dm}}{10 \mathrm{cm} \times 10 \mathrm{cm} \times 10 \mathrm{cm}} \)
\( \Rightarrow 1 \mathrm{mL}=10^{-3} \mathrm{dm}^{3} \)
\( \therefore 1 \mathrm{mL}=10^{-3} \mathrm{L}=10^{-3} \mathrm{dm}^{3} \) 

Ans : According to the question:
Time taken to cover the distance = 2.00 ns
\( =2.00 \times 10^{-9} \mathrm{s} \)
Speed of light \( =3.0 \times 10^{8} \mathrm{ms}^{-1} \)
Distance travelled by light in 2.00 ns
= Speed of light x Time taken
\( \begin{array}{l}{=\left(3.0 \times 10^{8} \mathrm{ms}^{-1}\right)\left(2.00 \times 10^{-9} \mathrm{s}\right)} \\ {=6.00 \times 10^{-1} \mathrm{m}} \\ {=0.600 \mathrm{m}}\end{array} \) 

Ans : A reagent determines the extent of a reaction. It is the reactant which is the first to get consumed during a reaction, thereby causing the reaction to stop and limiting the amount of products formed.
(i) According to the given reaction, 1 atom of A reacts with 1 molecule of B. Thus, 200 molecules of B will react with 200 atoms of A, thereby leaving 100 atoms of A unused. Hence, B is the Iimiting reagent.
(ii) According to the reaction, 1 mol of A reacts with 1 mol of 3. Thus, 2 mol of A will react with only 2 mol of B. As a result, 1 mol of A will not be consumed. Hence, A is the limiting reagent.
(iii) According to the given reaction, 1 atom of A combines with 1 molecule of B. Thus, all 100 atoms of A will combine with all 100 molecules of B. Hence, the mixture is stoichiometric where no limiting reagent is present.
(iv) 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of B will combine with only 2.5 mol of A. As a result, 2.5 mol of A will be left as such. Hence, 3 is the limiting reagent.
(v) According to the reaction, 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of A will combine with only 2.5 mol of B and the remaining 2.5 mol of B will be left as such. Hence, A is the limiting reagent. 

Ans : (i) Balancing the given chemical equation,
\( \mathrm{N}_{2(\mathrm{g})}+3 \mathrm{H}_{2(\mathrm{g})} \longrightarrow 2 \mathrm{NH}_{3(g)} \)
From equation, 1 mole (28 g) of nitrogen reacts with 3 mole (6 g) of dihydrogen to give 2 mole ( 34 g) of ammonia.
\( \Rightarrow 2.00 \times 10^{3} \mathrm{g} \) of di nitrogen will react with \( \frac{6 \mathrm{g}}{28 \mathrm{g}} \times 2.00 \times 10^{3} \mathrm{g} \) dihydrogen i.e., \( 2.00 \times 10^{3} \mathrm{g} \) of nitrogen will react with 428.6 of dihydrogen.
Given,
Amount of dihydrogen = \( 1.00 \times 10^{3} \mathrm{g} \)
Hence, \( \mathrm{N}_{2} \) is the limiting reagent.
\( \therefore 28 \mathrm{g} \text { of } \mathrm{N}_{2} \text { produces } 34 \mathrm{g} \text { of } \mathrm{NH}_{3} \)
Hence, mass of ammonia produced by 2000 g of \( \mathrm{N}_{2} \) \( =\frac{34 \mathrm{g}}{28 \mathrm{g}} \times 2000 \mathrm{g} \)
\( =2428.57 \mathrm{g} \)
(ii) \( \mathrm{N}_{2} \) is the limiting reagent and \( \mathrm{H}_{2} \) is the excess reagent. Hence, \( \mathrm{H}_{2} \) will remain unreacted.
(iii) Mass of dihydrogen left unreacted = \( 1.00 \times 10^{3} \mathrm{g}-428.6 \mathrm{g} \)
\( =571.4 \mathrm{g} \) 

Ans : Molar mass of \( \mathrm{Na}_{2} \mathrm{CO}_{3}=(2 \times 23)+12.00+(3 \times 16) \)
\( =106 \mathrm{g} \mathrm{mol}^{-1} \)
Now, 1 mole of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \text { means } 106 \mathrm{g} \text { of } \mathrm{Na}_{2} \mathrm{CO}_{3} \)
\(  \therefore 0.5 \mathrm{mol} \text { of } \mathrm{Na}_{2} \mathrm{CO}_{3} \) \( =\frac{106 \mathrm{g}}{1 \mathrm{mole}} \times 0.5 \mathrm{mol} \mathrm{Na}_{2} \mathrm{CO}_{3} \)
\( =53 \mathrm{g} \mathrm{Na}_{2} \mathrm{CO}_{3} \)
\( \Rightarrow 0.50 \mathrm{M} \text { of } \mathrm{Na}_{2} \mathrm{CO}_{3}=0.50 \mathrm{mol} / \mathrm{L} \mathrm{Na}_{2} \mathrm{CO}_{3} \)
Hence, \( 0.50 \mathrm{mol} \text { of } \mathrm{Na}_{2} \mathrm{CO}_{3} \) is present in 1 L of water or 53 g of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) is present in 1 L of water. 

Ans : Reaction of dihydrogen with dioxygen can be written as:
\( 2 \mathrm{H}_{2(\mathrm{s})}+\mathrm{O}_{2(x)} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}_{(x)} \)
Now, two volumes of dihydrogen react with one volume of dihydrogen to produce two volumes of water vapour. Hence, ten volumes of dihydrogen will react with five volumes of dioxygen to produce ten volumes of water vapour. 

Ans : (i) \( 28.7 \mathrm{pm} : \)
\( 1 \mathrm{pm}=10^{-12} \mathrm{m} \)
\( \begin{array}{l}{\therefore 28.7 \mathrm{pm}=28.7 \times 10^{-12} \mathrm{m}} \\ {=2.87 \times 10^{-11} \mathrm{m}}\end{array} \)
(ii) \( 15.15 \mathrm{pm} : \)
\( 1 \mathrm{pm}=10^{-12} \mathrm{m} \)
\( \begin{array}{l}{\therefore 15.15 \mathrm{pm}=15.15 \times 10^{-12} \mathrm{m}} \\ {=1.515 \times 10^{-12} \mathrm{m}}\end{array} \)
(iii) \( 25365 \mathrm{mg} \) :
\( 1 \mathrm{mg}=10^{-3} \mathrm{g} \)
\( 25365 \mathrm{mg}=2.5365 \times 10^{4} \times 10^{-3} \mathrm{g} \)
Since,
\( 1 \mathrm{g}=10^{-3} \mathrm{kg} \)
\( \begin{array}{l}{2.5365 \times 10^{1} \mathrm{g}=2.5365 \times 10^{-1} \times 10^{-3} \mathrm{kg}} \\ {\therefore 25365 \mathrm{mg}=2.5365 \times 10^{-2} \mathrm{kg}}\end{array} \) 

Ans : \( 1 \mathrm{g} \text { of } \mathrm{Au}(\mathrm{s}) \) \( =\frac{1}{197} \text { mol of } \mathrm{Au}(\mathrm{s}) \)
\( \begin{aligned} &=\frac{6.022 \times 10^{23}}{197} \text { atoms of Au (s) } \\ &=3.06 \times 10^{21} \text { atoms of } \mathrm{Au}(\mathrm{s}) \end{aligned} \)
\( 1 \mathrm{g} \text { of } \mathrm{Na}(\mathrm{s}) \) \( =\frac{1}{23} \mathrm{mol} \text { of } \mathrm{Na}(\mathrm{s}) \)
\( =\frac{6.022 \times 10^{23}}{23} \text { atoms of } \mathrm{Na}(\mathrm{s}) \)
\( \begin{array}{l}{=0.262 \times 10^{23} \text { atoms of } \mathrm{Na}(\mathrm{s})} \\ {=26.2 \times 10^{21} \text { atoms of } \mathrm{Na}(\mathrm{s})}\end{array} \)
\( 1 \text { g of } L(s) \) \( =\frac{1}{7} \mathrm{mol} \text { of } \mathrm{Li}(\mathrm{s}) \)
\( =\frac{6.022 \times 10^{33}}{7} \) \( atoms of Li (s) \)
\( \begin{array}{l}{=0.86 \times 10^{23} \text { atoms of Li (s) }} \\ {=86.0 \times 10^{21} \text { atoms of Li (s) }}\end{array} \)
\( 1 \mathrm{g} \text { of } \mathrm{Cl}_{2}(\mathrm{g}) \) \( =\frac{1}{71} \mathrm{mol} \text { of } \mathrm{Cl}_{2}(\mathrm{g}) \)
( Molar mass of \( \mathrm{Cl}_{2} \) molecule = \( 35.5 \times 2=71 \mathrm{g} \mathrm{mol}^{-1} \) )
\( =\frac{6.022 \times 10^{33}}{71} \text { atoms of } \mathrm{Cl}_{2}(\mathrm{g}) \)
\( \begin{array}{l}{=0.0848 \times 10^{23} \text { atoms of } \mathrm{Cl}_{2}(\mathrm{g})} \\ {=8.48 \times 10^{21} \text { atoms of } \mathrm{Cl}_{2}(\mathrm{g})}\end{array} \)
\( Hence, 1 g of Li (s) will have the largest number of atoms. \) 

Ans : Mole fraction of \( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \) \( =\frac{\text { Number of moles of } \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}{\text { Number of moles of solution }} \)

Number of moles present in 1 L water:
\( n_{\mathrm{H}_{2}O}=\frac{1000 \mathrm{g}}{18 \mathrm{g} \mathrm{mol}^{-1}} \)
\( n_{\mathrm{H}_{2} \mathrm{O}}=55.55 \mathrm{mol} \)
Substituting the value of \( n_{\mathbf{H}_{2} \mathrm{O}} \) in equation (1),
\( \frac{n_{\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}}}{n_{\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}}+55.55} \) \( =0.040 \)
\( n_{\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}}= \) \( 0.040 n_{\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}} \) + \( (0.040)(55.55) \)
\( 0.96 n_{\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}}= \) \( 2.222 \mathrm{mol} \)
\( h_{\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}} \) \( =\frac{2.222}{0.96} \mathrm{mol} \)
\( h_{\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}} \) \( \equiv 2.314 \mathrm{mol} \)
\( \therefore \) Molarity of solution = \( \frac{2.314 \mathrm{mol}}{1 \mathrm{L}} \)
\( =2.314 \mathrm{M} \) 

Ans : \( 1 \text { mole of carbon atoms }= \) \( 6.023 \times 10^{23} \) atoms of carbon
\( =12 \mathrm{g} \) of carbon
\( \therefore \text { Mass of one }^{12} \mathrm{C} \) = \( \frac{12 \mathrm{g}}{6.022 \times 10^{23}} \)
\( =1.993 \times 10^{-23} \mathrm{g} \) 

Ans : (i) \( \frac{0.02856 \times 298.15 \times 0.112}{0.5785} \)
Least precise number of calculation = 0.112
\( \therefore \) Number of significant figures in the answer.
= Number of significant figures in the least precise number.
= 3
(ii) \( 5 \times 5.364 \)
Least precise number of calculation \( =5.364 \)
\( \therefore \) Number of significant figures in the answer =  Number of significant figures in 5. 364
= 4
(iii) \( 0.0125+0.7864+0.0215 \)
Since the least number of decimal places in each term is four, the number of significant figures in the answer is also 4. 

Ans : Molar mass of argon
\( =\left[\left(35.96755 \times \frac{0.337}{100}\right)+\left(37.96272 \times \frac{0.063}{100}\right)+\left(39.9624 \times \frac{90.60}{100}\right)\right] \mathrm{gmol}^{-1} \)
\( \begin{aligned} &=[0.121+0.024+39.802] \mathrm{gmol}^{-1} \\ &=39.947 \mathrm{gmol}^{-1} \end{aligned} \) 

Ans : (i) \( 1 \text { mole of } \mathrm{Ar}=6.022 \times 10^{23} \text { atoms of } \mathrm{Ar} \)
\( \therefore 52 \text { mol of } \mathrm{Ar}=52 \times 6.022 \times 10^{23} \text { atoms of } \mathrm{Ar} \)
\( =3.131 \times 10^{25} \text { atoms of } \mathrm{Ar} \)
(ii) \( 1 \text { atom of He }= \) 4 u of He
Or,
\( 4 \mathrm{u} \text { of } \mathrm{He}=1 \) atom of He
\( 1 \mathrm{u} \text { of } \mathrm{He} \) \( =\frac{1}{4} \) atom of He
\( 52 \mathrm{u} \text { of } \mathrm{He} \) = \( \frac{52}{4} \) atom of He
\( =13 \) atoms of He
(iii) \( 4 \mathrm{g} \text { of } \mathrm{He}=6.022 \times 10^{23} \) atoms of He
\( \therefore \) 52 g of He = \( \frac{6.022 \times 10^{23} \times 52}{4} \) atoms of He
\( =7.8286 \times 10^{24} \) atoms of He 

Ans : (i) \( 1 \text { mole }\left(44 \text { g) of } \mathrm{CO}_{2} \text { contains } 12 \mathrm{g} \text { of carbon. }\right. \)
\( \therefore 3.38 \mathrm{g} \text { of } \mathrm{CO}_{2} \) will contain carbon \( =\frac{12 \mathrm{g}}{44 \mathrm{g}} \times 3.38 \mathrm{g} \)
\( =0.9217 g \)
\( 18 \text { g of water contains } 2 \) g of hydrogen.
\( \therefore 0.690 \) g of water will contain hydrogen = \( \frac{2 \mathrm{g}}{18 \mathrm{g}} \times 0.690 \)
\( =0.0767 \mathrm{g} \)
Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is:
\( \begin{aligned} &=0.9217 \mathrm{g}+0.0767 \mathrm{g} \\ &=0.9984 \mathrm{g} \end{aligned} \)
\( \therefore \text { Percent of } \mathrm{C} \) in the compound \( =\frac{0.9217 \mathrm{g}}{0.9984 \mathrm{g}} \times 100 \)
\( =92.32 \% \)
Percent of H in the compound \( =\frac{0.0767 \mathrm{g}}{0.9984 \mathrm{g}} \times 100 \)
\( =7.68 \% \)
Moles of carbon in the compound \( =\frac{92.32}{12.00} \)
\( =7.69 \)
Moles of hydrogen in the compound \( =\frac{7.68}{1} \)
\( =7.68 \)
\( \begin{array}{l}{\therefore \text { Ratio of carbon to hydrogen in the compound }=7.69 : 7.68} \\ {=1 : 1}\end{array} \)
Hence, the empirical formula of the gas is CH.

(ii) Given,
Weight of \( 10.0 \mathrm{L} \text { of the gas (at S.T.P) }=11.6 \mathrm{g} \)
\( \therefore \text { Weight of } 22.4 \) L of gas at STP = \( \frac{11.6 \mathrm{g}}{10.0 \mathrm{L}} \times 22.4 \mathrm{L} \)
\( \begin{array}{l}{=25.984 \mathrm{g}} \\ {\approx 26 \mathrm{g}}\end{array} \)
Hence, the molar mass of the gas is 26 g.

(iii) Empirical formula mass of CH = 12 + 1 = 13 g
\( n=\frac{\text { Molar mass of gas }}{\text { Empirical formula mass of gas }} \)
\( =\frac{26 \mathrm{g}}{13 \mathrm{g}} \)
\( n=2 \)
\( \begin{aligned} \therefore & \text { Molecular formula of gas }=(\mathrm{CH})_{n} \\ &=\mathrm{C}_{2} \mathrm{H}_{2} \end{aligned} \) 

Ans : \( 0.75 \mathrm{M} \text { of } \mathrm{HCl} \equiv 0.75 \mathrm{mol} \text { of } \mathrm{HCl} \text { are present in } 1 \mathrm{L} \) of water
\( \equiv\left[\left(0.75 \text { mol) } \times\left(36.5 \mathrm{g} \mathrm{mol}^{-1}\right)\right] \mathrm{HCl} \text { is present in } 1 \mathrm{L} \text { of water }\right. \)
\( \equiv 27.375 \mathrm{g} \text { of HCl is present in } 1 \mathrm{L} \) of water
Thus, 1000 mL of solution contains 27.375 g of HCL.
\( \therefore \text { Amount of HCl present in } 25 \mathrm{mL} \)
\( =\frac{27.375 \mathrm{g}}{1000 \mathrm{mL}} \times 25 \mathrm{mL} \)
\( =0.6844 \mathrm{g} \)
From the given chemical equation,
\( \mathrm{CaCO}_{3(s)}+2 \mathrm{HCl}_{(\mathrm{aq})} \longrightarrow \mathrm{CaCl}_{2(aq)} + \mathrm{CO}_{2(\mathrm{g})} +\mathrm{H}_{2} \mathrm{O}_{(g)} \)
\( 2 \text { mol of } \mathrm{HCl}(2 \times 36.5=71 \mathrm{g}) \) react with 1 mol of \( \mathrm{CaCO}_{3}(100 \mathrm{g}) \)
\( \therefore \text { Amount of } \mathrm{CaCO}_{3} \) that will react with \( 0.6844 \mathrm{g} \) \( =\frac{100}{71} \times 0.6844 \mathrm{g} \)
\( =0.9639 \mathrm{g} \) 

Ans : 1 mol \( [55+2 \times 16=87 \mathrm{g}] \mathrm{MnO}_{2} \) reacts completely with 4 mol \( [4 \times 36.5=146 \mathrm{g}] \) of HCL.
\( \therefore 5.0 \mathrm{g} \text { of MnO }_{2} \) will react with
\( =\frac{146 \mathrm{g}}{87 \mathrm{g}} \times 5.0 \mathrm{g} \) of HCL
\( =8.4 \mathrm{g} \text { of } \mathrm{HCl} \)
Hence, \( 8.4 \mathrm{g} \text { of } \mathrm{HCl} \) will react completely with \( 5.0 \mathrm{g} \) of manganese dioxide.