Question

The density of a mixture of $CO$ and $C{O}_2$ at $303\text{K}$ and $73\text{cm}$ $Hg$ is $1.5{\text{g L}}^{-1}$. What is the mole percent of $CO$ and $C{O}_2$ respectively in the mixture?

• A. $32.19\%,67.81\%$
• B. $31.19\%,67.81\%$
• C. $41.2\%,58.8\%$
• D. $40.2\%,58.8\%$

Answer 1

The correct option is A $32.19\%,67.81\%$

Let $M$ be the average molecular mass of the mixture.
$PV=\frac{wRT}{M}$
or, $P=\frac{wRT}{M\times V}$
or, $P=\frac{dRT}{M}$

$P=\frac{73}{76}\text{atm},d=1.5{\text{g L}}^{-1},M=?$
$T=303\text{K},R=0.0821{\text{L atm K}}^{-1}{\text{mol}}^{-1}$

$\therefore \frac{73}{76}=\frac{1.5\times 0.0821\times 303}{M}$
$M=38.85g/mol$

Suppose, the total moles = 100
Moles of $CO=a$
Moles 0f $C{O}_2=100-a$
Mass of $CO=28a$
Mass of $C{O}_2=(100-a)44$
Total mass $=28a+(100-a)44$
$\text{Molar mass of the mixture}=\frac{28a+(100-a)44}{100}$
$38.85=\frac{28a+(100-a)44}{100}$
$a=32.19$
$\text{Mole percent of}CO=32.19\%$
$\text{Mole percent of}C{O}_2=(100-32.19)\%=67.81\%$