5 g of Na_2SO_4 was dissolved in x g of H_2O . The change in freezing point was found to be 3.82oC. If Na_2SO_4 is 81.5% ionised, the value of x (Kf for water=1.86oC kg mol−1) is approximately : (molar mass of S = 32 g mol−1 and that of Na = 23 g mol−1)

Solutions PYQ — JEE Main 2017 · 8 Apr · Shift 1 · Q21 with Solution

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Correct answer: C

For a solute that ionises,

\Delta T_f = i K_f m

where:

M = 2(23) + 32 + 4(16) = 46 + 32 + 64 = 142\ g\,mol^{-1}

So, moles of $Na_2SO_4$ in $5\ g$ are:

n = \frac{5}{142}\ mol

$Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-}$

If fully ionised, total ions $=3$ . Given degree of ionisation $\alpha = 81.5\% = 0.815$ .

Hence,

i = 1 + (3-1)\alpha = 1 + 2(0.815) = 2.63

m = \frac{\Delta T_f}{iK_f} = \frac{3.82}{2.63 \times 1.86}

2.63 \times 1.86 = 4.8918

m \approx \frac{3.82}{4.8918} \approx 0.781\ mol\,kg^{-1}

Molality is:

m = \frac{\text{moles of solute}}{\text{kg of solvent}}

So,

0.781 = \frac{5/142}{x/1000}

\frac{5}{142} \approx 0.03521

Thus,

0.781 = \frac{0.03521}{x/1000}

x/1000 = \frac{0.03521}{0.781} \approx 0.0451

x \approx 45.1\ g

The nearest option is:

\boxed{45\ g}

So, Option C is correct.

Stored correct answer: C

Derived answer: C

They agree.