Question

5.8g of a non-volatile solute was dissolved in 100 g of carbon
disulphide $\left( {{\text{C}}{{\text{S}}_{\text{2}}}} \right)$. The vapour pressure of the solution was found to be $190{\text{ mm of Hg}}$. Calculate molar mass of solute given the vapour pressure of $\left( {{\text{C}}{{\text{S}}_{\text{2}}}} \right)$ is $195{\text{ mm of Hg}}$. [Molar mass of $\left( {{\text{C}}{{\text{S}}_{\text{2}}}} \right)$ $ = 76{\text{ g mo}}{{\text{l}}^{ - 1}}$]

Answer 1

Hint: To solve this we must know the Raoult’s law. We can calculate the molar mass of the solute using the formula for Raoult’s law. As the value of vapor pressure is given so we have to just put the value of it in the equation and just find out the value of mass.

Complete step by step solution:
We know Raoult's law. The Raoult’s law states that the partial vapour pressure of a solvent in a solution is equal to the vapour pressure of the pure solvent measured in the solution by the mole fraction.
The expression for Raoult’s law is as follows:
$\dfrac{{{p^0} - p}}{{{p^0}}} = \dfrac{{w \times M}}{{m \times W}}$
Where ${p^0}$ is the vapour pressure of the pure solvent,
$p$ is the vapour pressure of the solution,
$w$ is the weight of the non-volatile solute,
$m$ is the molar mass of the non-volatile solute,
$M$ is the molar mass of the pure solvent,
$W$ is the weight of the pure solvent.
We are given the following data:
Vapour pressure of pure solvent i.e. carbon disulphide $\left( {{\text{C}}{{\text{S}}_{\text{2}}}} \right)$ $ = 195{\text{ mm of Hg}}$.
Vapour pressure of solution $ = 190{\text{ mm of Hg}}$.
Weight of a non-volatile solute $ = 5.8{\text{ g}}$.
Weight of solvent i.e. carbon disulphide $\left( {{\text{C}}{{\text{S}}_{\text{2}}}} \right)$ $ = 100{\text{ g}}$.
Molar mass of the pure solvent i.e. carbon disulphide $\left( {{\text{C}}{{\text{S}}_{\text{2}}}} \right)$ $ = 76{\text{ g/mol}}$.
Substitute the given data in the expression for Raoult’s law. Thus,
$\dfrac{{\left( {195 - 190} \right){\text{mm of Hg}}}}{{195{\text{ mm of Hg}}}} = \dfrac{{5.8{\text{ g}} \times 76{\text{ g/mol}}}}{{m \times 100{\text{ g}}}}$
$\Rightarrow \dfrac{{5{\text{ mm of Hg}}}}{{195{\text{ mm of Hg}}}} = \dfrac{{4.408{\text{ g/mol}}}}{m}$
$\Rightarrow m = \dfrac{{4.408{\text{ g/mol}}}}{{0.02564}}$
$\Rightarrow m = 171.92{\text{ g/mol}}$

Thus, the molar mass of the solute is $171.92{\text{ g/mol}}$.

Note: We know that a solution is formed when we dissolve one substance into another substance. The substance dissolved is known as the solute and the dissolving material is known as the solvent. A solution is a homogeneous mixture which consists of solute dissolved in solvent. The molar mass is the total mass of all the atoms in the molecule. The units of molar mass are ${\text{g/mol}}$.