Question

The times taken for the effusion of equal volumes of carbon dioxide and a mixture of carbon dioxide with carbon monoxide were 28 s and 24 s respectively. Calculate the apparent molar mass of the mixture and the mole fraction of each gas in the mixture :

• A. $32.3g/mole,X_{C{O}_2}=0.27,X_{CO}=0.73$
• B. $12.4g/mole,X_{C{O}_2}=0.73,X_{CO}=0.27$
• C. $64.2g/mole,X_{C{O}_2}=0.18,X_{CO}=0.82$
• D. $22.3g/mole,X_{C{O}_2}=0.82,X_{CO}=0.18$

Answer 1

The correct option is A $32.3g/mole,X_{C{O}_2}=0.27,X_{CO}=0.73$

$Rateofeffusion=\frac{Volumeout}{Timetaken}$

$r_{C{O}_2}=\frac{Volumeout}{28}\frac{L}{S}$

$r_{mix}=\frac{Volumeout}{24}\frac{L}{S}$

Since we have taken equal volumes of $C{O}_2$ and mixtures of $C{O}_2+CO;$ Volume out will be same.

$\therefore r_{C{O}_2}\times 28=r_{mix}\times 24$

$⟹\frac{r_{C{O}_2}}{r_{mix}}=\frac{24}{28}=\frac{6}{7}$

From Graham's Law;

$\frac{r_2}{r_1}=\sqrt{\frac{M_1}{M_2}}$

Where $M_1$ and $M_2$ are molar mass

$⟹\frac{r_{C{O}_2}}{r_{mix}}=\sqrt{\frac{M_{mix}}{M_{C{O}_2}}}$

$⟹\frac{6}{7}=\sqrt{\frac{M_{mix}}{44}}$

$⟹M_{mix}=\frac{36}{49}\times 44=32.3$ $g/mol$

Let mole fraction of $C{O}_2$ in mixture be $x$.

$\therefore$ Mole fraction of $CO=1-x$

$Molarmassofmix=\frac{Molefraction\times M_{C{O}_2}+Molefraction\times M_{CO}}{MolefractionofC{O}_2+MolefractionofCO}$

$⟹32.3=\frac{x\times 44+(1-x)\times 28}{x+1-x}$

$⟹x=0.27$;

Mole fraction of $C{O}_2=0.27$

Mole fraction of $CO=1-0.27=0.73$