Question

The vapour pressure of a solution having $2.0\, g$ of solute $X$ (molar atomic mass $=32\, g\, mol ^{-1}$ ) in $100 \,g$ of $CS _{2}$ (vapour) pressure $=854$ torr) is $848.9$ torr. The molecular formula of the solute is

• A. $X$
• B. $X _{2}$
• C. $X _{4}$
• D. $X _{8}$

Answer 1

Answer: (D)

$\frac{p^{\circ}-p}{p^{\circ}}=\underset{(\text { Solute })}{x_{A}}, $
$x_{A}=\frac{n_{A}}{n_{A}+n_{B}} \simeq \frac{n_{A}}{n_{B}}$ (For dilute solution)
$\Rightarrow \frac{854-848.9}{854}=\frac{ n _{ A }}{ n _{ B }}=\frac{ W _{ A } \times M _{ B }}{ M _{ A } \times W _{ B }} ; $
$W _{ A }=$ weight of solute,
$W _{ B }=$ molar mass of solute
$\Rightarrow \frac{5.1}{854}=\frac{2 \times 76}{ M _{ A } \times 100} ;$
$ M _{ B }=$ molar mass of solvent
$= CS _{2}=12+32 \times 2$
$ \Rightarrow 76 \,g$
$\underset{\text{(Molar mass of solute)}}{M_A}=\frac{2 \times 76 \times 854}{100 \times 5.1}=253$
Number of atoms $=\frac{\text { molar mass }}{\text { atomic mass }}=\frac{253}{32}=7.90 \approx 8$
So formula of solute is $X _{8}$