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 the number of moles of glucose.
Given mass of glucose is \[ 1.8\,\text{g} \] Molar mass of glucose is \[ 180\,\text{g mol}^{-1} \] Therefore, \[ \text{Moles of glucose}=\frac{1.8}{180} \] \[ =0.01\,\text{mol} \]

Step 2: Calculate molality of the solution.
Molality is given by \[ m=\frac{\text{moles of solute}}{\text{mass of solvent in kg}} \] Given mass of water is \[ 0.1\,\text{kg} \] Therefore, \[ m=\frac{0.01}{0.1} \] \[ m=0.1\,\text{mol kg}^{-1} \]

Step 3: Calculate depression in freezing point.
For a non-electrolyte solute like glucose, \[ \Delta T_f=K_f m \] Given, \[ K_f=1.86\,\text{K kg mol}^{-1} \] So, \[ \Delta T_f=1.86\times 0.1 \] \[ \Delta T_f=0.186\,\text{K} \]

Step 4: Find freezing point of the solution.
The freezing point of pure water is \[ 0^\circ\text{C} \] Since freezing point is depressed, the freezing point of the solution is \[ 0-0.186 \] \[ =-0.186^\circ\text{C} \]

Step 5: Final conclusion.
Therefore, the freezing point of the solution is \[ \boxed{-0.186^\circ\text{C}} \] Hence, the correct option is \[ \boxed{(3)} \]