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CBSE Class 12 · 2016 · SolutionsShort Answer

Calculate the freezing point of a solution containing 60 g of glucose (molar mass = 180 g mol

^{- 1}

) in 250 g of water. (K

_{f}

of water = 1.86 K kg mol

^{- 1}

)

Answer: T_f ≈ −2.48°C (270.67 K)

Solution

Molality:

m = \frac{60/180}{0.250} = \frac{0.333}{0.250} = 1.333 mol kg^{- 1}

Depression of freezing point:

Δ T_{f} = K_{f} m = 1.86 \times 1.333 = 2.48 K

T_{f} = 0 - 2.48 = - 2.48^{\circ} C (= 270.67 K)

CBSE Class 12 Board · 2016 · Subjective

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