5 g of Na2SO4 was dissolved in x g of H2O. The change in freezing point was found to be 3.82°C. If Na2SO4 is 81.5% ionised, the value of x (Kf for water =1.86°Ckgmol-1) is approximately: ( molar mass of S=32gmol-1 and that of Na=36g mol m-1)

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5 g of ${Na}_{2} {SO}_{4}$ was dissolved in x g of $H_{2} O$ . The change in freezing point was found to be 3.82°C. If $N a_{2} S O_{4}$ is 81.5% ionised, the value of x ( $K_{f} for water = 1.86 ° {Ckgmol}^{- 1}$ ) is approximately: $(molar mass of S = 32 {gmol}^{- 1} a n d t h a t o f N a = 36 g m o l m^{- 1})$

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45g

15g

25g

65g

answer is A.

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Detailed Solution

$Molar mass of {Na}_{2} {SO}_{4} = 2 (23) + 32 + 4 (16) = 142 g / mol 5 {gNa}_{2} {SO}_{4} = \frac{5 g}{142 g / mol} = 0.0352 mol Mass of water = xg = \frac{x}{1000} kg = 0.001 xkg Molality m = \frac{0.0352 {molNa}_{2} {SO}_{4}}{0.001 x} = \frac{35.2}{x} molkg {Na}_{2} {SO}_{4} is 81.5 % ionised. degree of dissociation α = \frac{81.5}{100} = 0.815 Vant Hoff factor i = [1 + (n - 1) α] i = [1 + (3 - 1) \times 0.815] = 2.63 The decrease in freezing point {ΔT}_{f} = {iK}_{f} m \begin{array}{l} 3.82 ° C = 2.63 \times 1.86 {}^{\circ}{CKg} / mol \times \frac{35.2}{x} mol / kg \\ x = 45 g \end{array}$