CBSE Class 12 · 2007 · All India · SolutionsShort Answer

A 0.1539 molal aqueous solution of cane sugar (molar mass = 342 g mol

^{- 1}

) has a freezing point of 271 K while the freezing point of pure water is 273.15 K. What will be the freezing point of an aqueous solution containing 5 g of glucose (molar mass = 180 g mol

^{- 1}

) per 100 g of solution?

Answer: T_f ≈ 269.07 K (*)

Solution

From the sugar solution, the effective constant:

K_{f} = \frac{Δ T _{f}}{m} = \frac{273.15 - 271}{0.1539} = \frac{2.15}{0.1539} = 13.97 K kg mol^{- 1}

Glucose solution: 5 g glucose per 100 g solution → 95 g water:

m = \frac{5/180}{0.095} = 0.2924 mol kg^{- 1}

Depression:

Δ T_{f} = 13.97 \times 0.2924 = 4.08 K

T_{f} = 273.15 - 4.08 = 269.07 K

(*) The K

_{f}

implied by the given data is used, as the question intends — not water's literature value.

CBSE Class 12 Board · 2007 · All India · Subjective

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