Question

State Raoult’s law.
The vapour pressure of pure benzene at a certain temperature is $0.850$ bar. A non-volatile, non-electrolyte solid weighing $0.5$g when added to $39.0$g of benzene (molar mass 78g ${\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$. Vapour pressure of the solution, then is $0.845$ bar. What is the molar mass of the solid substance?

Answer 1

Raoult's law has implications in the thermodynamics. It is a law of physical chemistry which relates vapour pressure to mole fraction. It is established by the French chemist François-Marie Raoult. The vapour pressure of the system can be calculated from the formula given below.

Formula used:
$\dfrac{{{{\text{P}}^{\text{o}}}{\text{ - }}{{\text{P}}_{\text{s}}}}}{{{{\text{P}}^{\text{o}}}}}{\text{ = }}\dfrac{{{{\text{n}}_{\text{A}}}}}{{{{\text{n}}_{\text{A}}}{\text{ + }}{{\text{n}}_B}}}$
where ${{\text{P}}^{\text{o}}}$ is the pressure of the component in the pure state. Here ${{\text{P}}^{\text{o}}}$ is the pressure of benzene in pure state. ${{\text{P}}_{\text{s}}}$ is the pressure of the solution.

Complete step-by-step answer: Raoult's law is given for volatile and non-volatile solutes. Raoult's law for volatile solute is given as follows; For volatile solute, the vapour pressure of any component at a given temperature is equal to the mole fraction of that component in the solution multiplied by the vapour pressure of that component in the pure state.
Raoult's law for non-volatile solute is given as follows; Relative lowering in vapour pressure of a solution containing a non-volatile solute is equal to the mole fraction of a non-volatile solute in the solution.
In the given question non-volatile solute is added and hence the formula used is as follows,
According to the Raoult’s law for non-volatile solute,
$\dfrac{{{{\text{P}}^{\text{o}}}{\text{ - }}{{\text{P}}_{\text{s}}}}}{{{{\text{P}}^{\text{o}}}}}{\text{ = }}\dfrac{{{{\text{n}}_{\text{A}}}}}{{{{\text{n}}_{\text{A}}}{\text{ + }}{{\text{n}}_B}}}$
where ${{\text{P}}^{\text{o}}}$ is the pressure of the component in the pure state. Here ${{\text{P}}^{\text{o}}}$ is the pressure of benzene in pure state. ${{\text{P}}_{\text{s}}}$ is the pressure of the solution.
The pressure of the benzene in the pure state is $0.850$ bar. The vapor pressure of the solution is $0.845$ bar.
${{\text{n}}_{\text{A}}}$ are the moles of non-volatile solute and ${{\text{n}}_B}$ are the moles of the solvent which is benzene here.
Let M be the molar mass of the solid substance. The given mass of non-volatile solute is $0.5$ g. Hence, ${{\text{n}}_{\text{A}}}{\text{ = }}\dfrac{{{\text{0}}{\text{.5}}}}{{\text{M}}}$
The given mass of benzene is 39 g and the molar mass of benzene is 78 g. Hence, ${{\text{n}}_B}{\text{ = }}\dfrac{{39}}{{78}}$
Hence the equation is
$\dfrac{{{\text{0}}{\text{.850 - 0}}{\text{.845}}}}{{{\text{0}}{\text{.850}}}}{\text{ = }}\left( {\dfrac{{\dfrac{{{\text{0}}{\text{.5}}}}{{\text{M}}}}}{{\dfrac{{{\text{0}}{\text{.5}}}}{{\text{M}}}{\text{ + 0}}{\text{.5}}}}} \right) = {\text{ }}\dfrac{{{\text{0}}{\text{.5}}}}{{{\text{0}}{\text{.5 + 0}}{\text{.5M}}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{1 + M}}}}$
${\text{M = }}\dfrac{{0.850}}{{0.005}}{\text{ - 1 = 169 g/mol }}$
Hence, the molar mass of the non-volatile solute is 169 g.

Note: Certain solutions do not obey Raoult's law. There can be positive deviation from raoult's law or negative deviation from raoult's law.