Question

\(1.8\,\text{g}\) of glucose \((\text{molar mass }=180\,\text{g mol}^{-1})\) is dissolved in \(0.1\,\text{kg}\) of water. The freezing point of the solution in \(^\circ\text{C}\) is \((K_f \text{ of water}=1.86\,\text{K kg mol}^{-1})\):

• A. \(+0.186\)
• B. \(-0.372\)
• C. \(-0.186\)
• D. \(+0.372\)

Hint:

For non-electrolyte solutions, \[ \Delta T_f=K_fm \] The actual freezing point of solution is lower than that of pure solvent: \[ T_f = 0^\circ\text{C}-\Delta T_f \] for aqueous solutions.

Answer 1

The Correct Option is C

Step 1: Calculate molality.
Moles of glucose = 1.8/180 = 0.01 mol. Mass of solvent = 0.1 kg. Molality \(m\) = 0.01/0.1 = 0.1 mol/kg.

Step 2: Apply depression in freezing point formula.
\(\Delta T_f = K_f imes m = 1.86 imes 0.1 = 0.186\) K. Since freezing point is depressed, \(T_f = 0 - 0.186 = -0.186\,^\circ\)C. \[ oxed{-0.186} \]