Question

1.8 g of glucose (molar mass 180 g/mol) is dissolved in 0.1 kg of water. The freezing point of the solution (in °C) is
(K$_f$ for water = 1.86 K kg mol$^{-1}$)

• A. -0.186
• B. -0.372
• C. -0.744
• D. -0.372

Hint:

Use $\Delta T_f = K_f \cdot m$ to calculate freezing point depression
Ensure units are correct for molality and K$_f$

Answer 1

The Correct Option is B

\[ \text{Moles of glucose} = \frac{1.8}{180} = 0.01 \text{ mol} \] \[ \text{Molality} = \frac{0.01}{0.1} = 0.1 \text{ mol/kg} \] \[ \Delta T_f = K_f \times m = 1.86 \times 0.2 = 0.372 \] \[ \text{Freezing point depression} = -0.372^\circ C \]