The freezing point is reduced from $5.51$ to $5.03^\circ\text{C}$ when $0.721\text{ g}$ of a compound is added to $75\text{ mL}$ of benzene. ($\text{density of benzene} = 0.879\text{ g/mL}$, $K_f \text{ for benzene} = 5.12\text{ K kg mol}^{-1}$). Calculate the molecular mass of the compound.

$116.65\text{ g/mol}$

The correct answer is Option (2) → $116.65\text{ g/mol}$ ##

Let Molecular mass of compound be A

$\text{Since, Density (D)} = \frac{\text{M}}{\text{V}}$

$0.879 = \frac{\text{Mass}}{75}$

$\text{Mass (benzene)} = 65.925\text{ g}$

$= 0.06593\text{ kg}$

$\text{Moles of solute} = \frac{\text{Wt of solute}}{\text{Molecular mass}}$

$= \frac{0.721}{\text{A}}$

$\text{Molality} = \frac{\text{Moles of solute}}{\text{Mass of benzene}}$

$= \frac{0.721}{\text{A} \times 0.06593}$

$\text{Depression in freezing point} = \Delta T_f = K_f \times m$

$5.51 - 5.03 = 5.12 \times \frac{0.721}{\text{A} \times 0.06593}$

$\text{A} = \frac{3.691}{0.0316}$

$\text{A} = 116.65\text{ g/mol}$