Question

The vapour pressure of a solution having $2.0\,g$of solute $X\,($gram atomic mass $ = \,32\,g\,mo{l^{ - 1}}\,)$in $100\,g$of $C{S_2}\,($vapour pressure $ = \,854\,torr\,)$ is $848.9\,torr$. The molecular formula of the solute is
(i) $X$
(ii) ${X_2}$
(iii) ${X_4}$
(iv) ${X_8}$

Answer 1

Hint: Consider the molecular formula of the solute to be ${X_m}$, hence the molecular weight of the solute will be equal to $\left( {32 \times m} \right)\,g\,mo{l^{ - 1}}$. From here calculate the number of moles of solute. Then use the equation $\dfrac{{\Delta P}}{{{P^ \circ }}} = \dfrac{{{n_2}}}{{{n_1}}}$ to find the value of $m$, where $2$ represents the solute and $1$ represents the solvent.

Complete step-by-step answer:In this problem we will consider $1$ as the solvent and $2$ as the solute.
Now, let the vapour pressure of the pure solvent be ${P^ \circ }$and the vapour pressure of the solution be $P$.
Then, the lowering in vapour pressure is given as, $\Delta P\, = \,{P^ \circ } - P$
So the relative lowering of vapour pressure is given by,
$\dfrac{{\Delta P}}{{{P^ \circ }}} = \dfrac{{{n_2}}}{{{n_1}}}$
$ \Rightarrow \dfrac{{{P^ \circ } - P}}{{{P^ \circ }}} = \dfrac{{{n_2}}}{{{n_1}}}.........\left( 1 \right)$
It is given that the vapour pressure of the pure solvent $C{S_2}$ i.e. ${P^ \circ }\, = \,854\,torr$and the vapour pressure of the solution i.e. $P\, = \,848.9\,torr$.
Now, ${n_1}$is the number of moles of $C{S_2}$ present in the solution.
We know, $No.\,of\,moles\,of\,a\,given\,compound\, = \,\dfrac{{Given\,Weight}}{{Molecular\,Weight}}$
Molecular Weight of $C{S_2}\, = \,76\,g\,mo{l^{ - 1}}$and the given weight of $C{S_2}\, = \,100\,g$.
Therefore, the number of moles of $C{S_2}$ i.e. ${n_1}\, = \,\dfrac{{100\,g}}{{76\,g\,mo{l^{ - 1}}}}\, = \,1.316\,mol$.
${n_2}$is the number of moles of solute present in the solution.
Let the molecular formula of the solute to be ${X_m}$, hence the molecular weight of the solute will be equal to $\left( {32 \times m} \right)\,g\,mo{l^{ - 1}}$ and the given weight of the solute $ = \,2\,g$.
Therefore, the number of moles of solute i.e. ${n_2}\, = \,\dfrac{{2\,g}}{{\left( {32 \times m} \right)\,g\,mo{l^{ - 1}}}}\, = \,\dfrac{1}{{16 \times m}}\,mol$.
Hence putting all the values in equation $\left( 1 \right)$ we get,
$\dfrac{{\left( {854 - 848.9} \right)\,torr}}{{854\,torr}} = \dfrac{{1\,mol}}{{\left( {16 \times m \times 1.316} \right)\,mol}}$
\[ \Rightarrow \,\dfrac{{5.1}}{{854}}\, = \,\dfrac{1}{{21.056 \times m}}\]
$ \Rightarrow \,0.00597\, = \,\dfrac{1}{{21.056 \times m}}$
$ \Rightarrow \,m\, = \,\dfrac{1}{{21.056 \times 0.00597}}\, = \dfrac{1}{{0.1257}}\, = \,7.955\, \sim \,8$
Hence the molecular formula of the solute is ${X_8}$.

So the correct answer is (iv) ${X_8}$.

Note: Students often make mistakes in the units in these kinds of questions. So in order to avoid mistakes try to do the question in a stepwise manner so that you know where the same units are cancelling out. Also do not get confused between the solute and the solvent and follow the same convention throughout.