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

The correct option is D none of these
Solution:-
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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