Question

Calculate the molality of nonvolatile solution if solution freezes at \(-0.95^{\circ} \mathrm{C}\)

\(\left[K_f\right.\) for water, \(=1.86 \mathrm{~K} \mathrm{~mol}^{-1}\), freezing point of water \(=0^{\circ} \mathrm{C}\) ]

• A. \(0.032\text{ }\mathrm{mol}\text{ }\mathrm{kg}^{-1}\)
• B. \(0.041\text{ }\mathrm{mol}\text{ }\mathrm{kg}^{-1}\)
• C. \(0.51\text{ }\mathrm{mol}\text{ }\mathrm{kg}^{-1}\)
• D. \(0.065\text{ }\mathrm{mol}\text{ }\mathrm{kg}^{-1}\)

Hint:

Use \(\Delta T_f = K_f \times m\) with \(\Delta T_f = 0 - (-0.95)\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The depression in freezing point is proportional to the molality of the solution: \(\Delta T_f = K_f\, m\).

Step 2: Data
\(\Delta T_f = 0^{\circ}\text{C} - (-0.95^{\circ}\text{C}) = 0.95\text{ K}\)
\(K_f = 1.86\text{ K kg mol}^{-1}\)

Step 3: Calculation
\(m = \dfrac{\Delta T_f}{K_f} = \dfrac{0.95}{1.86} = 0.51\text{ mol kg}^{-1}\)

Final Answer:
The molality of the solution is 0.51 mol per kg. \[ \boxed{\text{(C) }0.51\text{ mol kg}^{-1}} \]