JEE Main 2017 · 8 Apr · Shift 1 · Chemistry Q37

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ChemistrySingle correct+4 / -1Question 37 of 90

5 g of $\ce{Na2SO4}$ was dissolved in $x$ g of $\ce{H2O}$ . The change in freezing point was found to be 3.82°C. If $\ce{Na2SO4}$ is 81.5% ionised, the value of $x$ ( $K_f$ for water = 1.86°C kg mol $^{-1}$ ) is approximately :

(molar mass of S = 32 g mol $^{-1}$ and that of Na = 23 g mol $^{-1}$ )

15 g

25 g

45 g

65 g

Answer and worked solution

Official answer: C

$i=1+2\alpha=2.63$ and $\ce{Na2SO4}$ is 142 g mol⁻¹:
$3.82=2.63\times1.86\times\frac{5/142}{x/1000}\implies x\approx45\text{ g}.$

Topics:van't Hoff factor Colligative properties

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JEE Main 2017 · 8 Apr · Shift 1, Chemistry question 37