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 = \( 82.0245 \mathrm{g} \mathrm{mole}^{-1}(\text { Given }) \)
\( \therefore \text { Required mass of sodium acetate }=\left(82.0245 \mathrm{g} \mathrm{mol}^{-1}\right)(0.1875 \text { mole) } \)
\( =15.38 \mathrm{g} \)