2 L of a mixture of nitrous oxide and nitric oxide at STP have a mean molecular weight of 39.8. When this mixture passed over red hot copper both dissociate to given nitrogen and oxygen gas. The total volume of Nitrogen and oxygen obtained at STP respectively are-

Engineering

Chemistry

Eudiometry

Question

1 L, 1.3 L

1 L, 1.7 L

1.3 L, 1L

1.7 L, 1 L

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M_avg =   $X_{N_{2} O} \cdot M_{N_{2} O} + X_{NO} \cdot M_{NO} = 39 .8$
Let $X_{N_{2} O}$ = x
$∴$ X_NO = 1 – x
x . 44 + (1– x) . 30 = 39.8
x = 0.7
⇒   $X_{N_{2} O}$ = 0.7
⇒ Volume fraction of N₂O = 0.7
⇒ Volume of N₂O = 2 × 0.7 = 1.4 L
⇒ Volume of NO = 2 – 1.4 = 0.6 L
Now,
       $\overset{1 .4 litre}{N_{2} O} \overset{}{\to} \underset{1 .4}{N_{2}} + \frac{1}{2} \underset{0 .7}{O_{2}}$
$\overset{0 .6}{NO} \overset{}{\to} \frac{1}{2} \underset{0 .3}{N_{2}} + \frac{1}{2} \underset{0 .3}{O_{2}}$
Total volume of N₂ = 1.4 + 0.3 = 1.7 L
Total volume of O₂ = 0.7 + 0.3 = 1 L