A solution of sucrose (molar mass = 342 g mol^-1 ) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the sol - Chemistry MCQ Chemistry | ExamTest.live | ExamTest.live

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A solution of sucrose (molar mass = 342 g mol^{- 1}) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be (K_{f} for water = 1.86 K kg mol^{- 1}):

A $- 0.57 0^{\circ} C$
B $- 0.37 2^{\circ} C$
C $- 0.52 0^{\circ} C$
D $+ 0.37 2^{\circ} C$

Solution & Step-by-step Explanation

1. Calculate Molality (m): Moles of sucrose = 68.5/342 = 0.2 mol m = \frac{0.2 mol}{1 kg solvent} = 0.2 m Calculate Depression in Freezing Point (Δ T_{f}): Δ T_{f} = K_{f} \cdot m Δ T_{f} = 1.86 \times 0.2 = 0.372 K (or 0.37 2^{\circ} C)Find Freezing Point of Solution: T_{f} = T_{f}^{\circ} - Δ T_{f} = 0 - 0.372 = - 0.37 2^{\circ} C

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A solution of sucrose (molar mass = 342 g mol^{- 1}) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be (K_{f} for water = 1.86 K kg mol^{- 1}):

A

- 0.57 0^{\circ} C

B

- 0.37 2^{\circ} C

C

- 0.52 0^{\circ} C

D

+ 0.37 2^{\circ} C

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Solutions Colligative Properties Freezing Point

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