Calculate the freezing point depression expected for a 0.05 m aqueous solution of Na$_2$SO$_4$. If this solution actually freezes at −0.320°C, what would be the value of van't Hoff factor? (K$_f$ for water is 1.86 K kg mol$^{-1}$)

CBSE Class 12 · 2009 · Delhi · SolutionsShort Answer

Calculate the freezing point depression expected for a 0.05 m aqueous solution of Na

_{2}

_{4}

. If this solution actually freezes at −0.320°C, what would be the value of van't Hoff factor? (K

_{f}

for water is 1.86 K kg mol

^{- 1}

)

Answer: ΔT_f(calc) = 0.093 K; i ≈ 3.44 (*)

Solution

Expected depression (no dissociation):

Δ T_{f}^{c a l c} = K_{f} m = 1.86 \times 0.05

Δ T_{f}^{c a l c} = 0.093 K

Observed depression = 0.320 K, so

i = \frac{0.320}{0.093}

i = 3.44

(*) The booklet's OCR prints the observed freezing point as −0.0320°C; the Delhi 2009 paper value is −0.320°C, which is used here. (The value i > 3 simply reflects the question's constructed data.)

CBSE Class 12 Board · 2009 · Delhi · Subjective

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