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Solutions

Question

Calculate the temperature at which a solution containing 54 g of glucose, C₃H₁₂O₆, in 250 g of water will freeze. [K_ffor water = 1.86 K kg mol^–1]

Solution

Answer:

by applying the formula

∆ T_{f} = \frac{K_{f} \times W_{B} \times 1000}{W_{A} \times M_{B}} W_{A} = 250 g, W_{B} = 54 g, M_{B} = 180 g K_{f} = 1.86 K kg {mol}^{- 1}

Substituting the values

∆ T_{f} = \frac{1.86 \times 54 \times 1000}{250 \times 180} = 2.232 K

∴

Freezing point of solution
= 273 - 2.232 = 270.768 K.