Question

# The air contains $$78%$$ of $$N_2$$ & $$22\%$$ of $${ O }_{ 2 }$$ by volume. The volume occupied by $$40 gm$$ air at $${ 20 }^{ o }C$$ & $$745$$ mm Hg pressure approximately:

A
34 lit
B
34 ml
C
3.4 K. lit.
D
none of these

Solution

## The correct option is D none of theseSolution:-Since the air contains $$78 \% \; {N}_{2}$$ and $$22 \% \; {O}_{2}$$ by volume.Therefore, given weight of air $$= 40 \text{ gm}$$Mass of $${N}_{2}$$ in $$40 \text{ gm}$$ of air $$= \cfrac{78}{100} \times 40 = 31.2 \text{ gm}$$Mass of $${O}_{2}$$ in $$40 \text{ gm}$$ of air $$= \cfrac{22}{100} \times 40 = 8.8 \text{ gm}$$Molar mass of $${N}_{2} = 28 \text{ gm}$$Molar mass of $${O}_{2} = 32 \text{ gm}$$As we know that,$$\text{No. of moles} = \cfrac{\text{mass}}{\text{molar mass}}$$$$\therefore \; \text{No. of moles in of } {N}_{2} \text{ in 40 gm of air} = \cfrac{31.2}{28} = 1.114 \text{ mole}$$$$\text{No. of moles in of } {O}_{2} \text{ in 40 gm of air} = \cfrac{8.8}{32} = 0.275 \text{ mole}$$As we know that volume occupied by 1 mole of a gas is $$22.4 L$$.Therefore,Volume occupied by 1.114 moles of $${N}_{2} = 22.4 \times 1.114 = 24.95 L$$Volume occupied by 0.275 moles of $${O}_{2} = 22.4 \times 0.275 = 6.16 L$$$$\therefore$$ Total volume occupied by $$40 \text{ gm}$$ of air $$= {V}_{{N}_{2}} + {V}_{{O}_{2}} = \left( 24.95 + 6.16 \right) L = 31.11 L$$Hence, the volume occupied by $$40 \text{ gm}$$ of air is $$31.11 L$$.Chemistry

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