Question

Q. An aqueous solution containing 12.5 x 10-3 kg of non-volatile compound in 0.1 kg of water freezes at 272.49 K. Determine molar mass of solute. [Kb for water = 1.86 K. kg mol-1, B.P. of water = 273.15 K]​

Answer 1

Answer:

Molar mass of the solute = 352.27 g mol⁻¹

Explanation:

Given:

Mass of the volatile compound (solute) = 12.5 × 10⁻³ kg = 12.5 g

Mass of water (solvent) = 0.1 kg = 100 g

Freezing point of the solution = 272.49 K

Cryoscopic constant of water = 1.86 K kg mol⁻¹

Freezing point of water = 273.15 K

To Find:

Molar mass of the solute

Solution:

Here we have to find the molar mass of the volatile compound.

Molar mass of the solute is given by the equation,

$\boxed{\sf M_2=\dfrac{K_f\times w_2\times 1000}{\Delta T_f\times w_1 }}$

where M₂ is the molar mass of the solute, $\sf K_f$ is the cryoscopic constant,

w₂ is the weight of the solute, w₁ is the weight of the solvent, $\sf \Delta T_f$ is the depression in freezing point.

$\sf \Delta T_F=273.15\:K-272.49\:K$

$\sf \implies 0.66\: K$

Now substitute the values in the above equation,

$\sf M_2=\dfrac{1.86\: K\:kg\:mol^{-1}\times 12.5\:g\times 1000\:g\:kg^{-1}}{0.66\:K\times 100\:g}$

$\sf M_2=\dfrac{232.5}{0.66}$

$\sf M_2=352.27\:g\:mol^{-1}$

Hence the molar mass of the solute is 352.27 g mol⁻¹