Question

The vapor pressure of pure water at $25^\circ {\rm{C}}$ is 23.76 torr. The vapor pressure of a solution containing 5.40 g of a non-volatile substance is 90.0 g water is 23.32 torr. The molecular weight of the solute is:
A. 97.24 g/mol
B. .24.29 g/mol
C. 50.44 g/mol
D. .57.24 g/mol

Answer 1

We know that mole fraction (solute) is equal to relative lowering of vapor pressure. Here, we have to use the formula $\dfrac{{p_1^0 - {p_1}}}{{p_1^0}} = {x_2}$, where, $p_1^0$ refers to vapor pressure (pure water), ${p_1}$ refers to vapor pressure of solution containing non-volatile solute and ${x_2}$ refers to mole fraction of solute.

Complete step by step answer: Vapor pressure of pure water is given as 23.76 torr and vapor pressure of solution containing non-volatile solute is given as 23.32 torr.

Now, we have to calculate the mole fraction of solute. To calculate mole fraction, we need the moles of solvent (water) and solute (non-volatile substance).

Let’s calculate moles of water first. Mass of water is given as 90.0 g and molar mass of water is ${{\rm{H}}_{\rm{2}}}{\rm{O}} = {\rm{2}} \times {\rm{1}}\;{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}{\rm{ + 16}}\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}} = 18\;{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$
So,
${\rm{Moles}}\;{\rm{of}}\;{\rm{water}} = \dfrac{{{\rm{Mass}}\;{\rm{of}}\;{\rm{water}}}}{{{\rm{Molar mass}}\;{\rm{of}}\;{\rm{water}}}}$
$\Rightarrow \dfrac{{90\,{\rm{g}}}}{{18\;{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}}}$
$\Rightarrow 5\;{\rm{mol}}$
Now, we have to calculate moles of solute (non-volatile). Mass of solute is given as 5.40 g and molar mass of solute is M.
${\rm{Moles}}\;{\rm{of}}\;{\rm{solute}} = \dfrac{{5.40\;{\rm{g}}}}{M}$
Now, we write the expression of mole fraction of solute.
${\rm{Mole}}\;{\rm{fraction}}\;{\rm{of}}\;{\rm{solute}} = \dfrac{{{\rm{Moles}}\;{\rm{of}}\;{\rm{solute}}}}{{{\rm{Moles}}\;{\rm{of}}\;{\rm{solute}} + {\rm{Moles}}\;{\rm{of}}\;{\rm{solvent}}}}$
$\Rightarrow \dfrac{{\dfrac{{5.40\;}}{M}}}{{\dfrac{{5.40\;}}{M} + 5}}$
$\Rightarrow \dfrac{{\dfrac{{5.40}}{M}}}{{\dfrac{{5.40 + 5M}}{M}}}$
$\Rightarrow \dfrac{{5.40}}{{5.40 + 5M}}$
Now, we have used the expression of lowering of vapour pressure to calculate the molecular mass of solute.
$\dfrac{{p_1^0 - {p_1}}}{{p_1^0}} = {x_2}$
…… (1)

From the question, $p_1^0$=23.76 torr and ${p_1} =$23.32 torr and mole fraction of solute $\left( {{x_2}} \right)$ is $\dfrac{{5.40}}{{5.40 + 5M}}$
Put all the above values in equation (1).
$\dfrac{{23.76\,{\rm{torr}} - 23.32\;{\rm{torr}}}}{{23.76\,{\rm{torr}}}} = \dfrac{{5.40}}{{5.40 + 5M}}$
$\Rightarrow \dfrac{{0.44}}{{23.76}} = \dfrac{{5.40}}{{5.40 + 5M}}$
$\Rightarrow \dfrac{{0.44}}{{23.76}} = \dfrac{{5.40}}{{5.40 + 5M}}$
$\Rightarrow 0.01852 = \dfrac{{5.40}}{{5.40 + 5M}}$
$\Rightarrow 5.40 + 5M = 291.6$
$\Rightarrow M = \dfrac{{291.6 - 5.40}}{5}$
$\Rightarrow M = 57.24\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}$

So, the correct answer is “Option D”.

Note: Colligative property is that property which depends on the count of particles of solute relative to the total count of particles present in a solution irrespective of their nature. Vapour pressure is an example of colligative property.