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.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $116.65\text{ g/mol}$ ##
Let Molecular mass of compound be A
$\text{Since, Density (D)} = \frac{\text{M}}{\text{V}}$
$0.879 = \frac{\text{Mass}}{75}$
$\text{Mass (benzene)} = 65.925\text{ g}$
$= 0.06593\text{ kg}$
$\text{Moles of solute} = \frac{\text{Wt of solute}}{\text{Molecular mass}}$
$= \frac{0.721}{\text{A}}$
$\text{Molality} = \frac{\text{Moles of solute}}{\text{Mass of benzene}}$
$= \frac{0.721}{\text{A} \times 0.06593}$
$\text{Depression in freezing point} = \Delta T_f = K_f \times m$
$5.51 - 5.03 = 5.12 \times \frac{0.721}{\text{A} \times 0.06593}$
$\text{A} = \frac{3.691}{0.0316}$
$\text{A} = 116.65\text{ g/mol}$