Question

\(45\,g\) urea \((CO(NH_2)_2)\) dissolved in \(1000\,g\) of water depresses the freezing point by \(1.395\,K\). The depression in freezing point of a solution of \(90\,g\) glucose \((C_6H_{12}O_6)\) in \(2000\,g\) water will be

• A. \(0.465\,K\)
• B. \(0.930\,K\)
• C. \(1.395\,K\)
• D. \(1.860\,K\)

Hint:

For non-electrolytes: \[ \Delta T_f \propto \text{Molality} \] If \(K_f\) and solvent are same, compare only molalities.

Answer 1

The Correct Option is A

Concept: Depression in freezing point is given by \[\begin{aligned} \Delta T_f=K_f m \end{aligned}\] where \(m\) is the molality. For non-electrolytes, \[\begin{aligned} \Delta T_f \propto m \end{aligned}\]

Step 1: Calculate molality of urea solution. Molar mass of urea \[\begin{aligned} M=60\,g\,mol^{-1} \end{aligned}\] Moles of urea \[\begin{aligned} \frac{45}{60}=0.75 \end{aligned}\] Molality \[\begin{aligned} m_1=\frac{0.75}{1}=0.75 \end{aligned}\]

Step 2: Calculate molality of glucose solution. Molar mass of glucose \[\begin{aligned} M=180\,g\,mol^{-1} \end{aligned}\] Moles of glucose \[\begin{aligned} \frac{90}{180}=0.5 \end{aligned}\] Mass of solvent \[\begin{aligned} 2000\,g=2\,kg \end{aligned}\] Molality \[\begin{aligned} m_2=\frac{0.5}{2}=0.25 \end{aligned}\]

Step 3: Use proportionality. \[\begin{aligned} \frac{\Delta T_{f2}}{\Delta T_{f1}} = \frac{m_2}{m_1} = \frac{0.25}{0.75} = \frac{1}{3} \end{aligned}\] Therefore, \[\begin{aligned} \Delta T_{f2} = 1.395\times\frac{1}{3} = 0.465\,K \end{aligned}\] \[\begin{aligned} \boxed{0.465\,K} \end{aligned}\] Hence, option \(\mathbf{(A)}\) is correct.