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CBSE Class 12 · 2008 · All India · SolutionsShort Answer

Calculate the temperature at which a solution containing 54 g of glucose (C

_{6}

H

_{12}

O

_{6}

) in 250 g of water will freeze. (K

_{f}

for water = 1.86 K kg mol

^{- 1}

)

Answer: T_f ≈ −2.23°C (270.92 K)

Solution

Molality:

m = \frac{54/180}{0.250} = \frac{0.3}{0.250} = 1.2 mol kg^{- 1}

Depression:

Δ T_{f} = 1.86 \times 1.2 = 2.232 K

T_{f} = 0 - 2.23 = - 2.23^{\circ} C (= 270.92 K)

CBSE Class 12 Board · 2008 · All India · Subjective

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