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CBSE Class 12 · 2016 · SolutionsShort Answer

Calculate the boiling point of a solution when 2 g of Na

_{2}

SO

_{4}

(M = 142 g mol

^{- 1}

) was dissolved in 50 g of water, assuming Na

_{2}

SO

_{4}

undergoes complete ionization. (K

_{b}

for water = 0.52 K kg mol

^{- 1}

)

Answer: T_b ≈ 100.44°C

Solution

Molality:

m = \frac{2/142}{0.050} = 0.2817 mol kg^{- 1}

Complete ionisation of Na

_{2}

SO

_{4}

(2Na

^{+}

+ SO

_{4}^{2 -}

) gives

i = 3

:

Δ T_{b} = i K_{b} m = 3 \times 0.52 \times 0.2817 = 0.44 K

T_{b} = 100 + 0.44 = 100.44^{\circ} C

CBSE Class 12 Board · 2016 · Subjective

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