CBSE Class 12 · 2002 · Delhi Compartment · SolutionsShort Answer

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In a solution of urea, 3.0 g of it is dissolved in 100 mL of water. What will be the freezing point of this solution? State the approximation made, if any. (K

_{f}

for water = 1.86 K kg mol

^{- 1}

, molar mass of urea = 60 g mol

^{- 1}

)

Answer: T_f ≈ −0.93°C. Approximation: 100 mL of water is taken as 100 g (density 1 g/mL) and the solution treated as dilute.

Solution

Taking the density of water as 1 g mL

^{- 1}

, mass of water = 100 g = 0.1 kg (the approximation made).
Molality:

m = \frac{3/60}{0.1} = 0.5 mol kg^{- 1}

Depression of freezing point:

Δ T_{f} = K_{f} m = 1.86 \times 0.5 = 0.93 K

T_{f} = 0 - 0.93 = - 0.93^{\circ} C

CBSE Class 12 Board · 2002 · Delhi Compartment · Subjective

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