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10 ml of $F e C_{2} O_{4}$ solution requires 30 ml of 0.4 M $K M n O_{4}$ solution in acidic medium for titration. 25 ml of the same $F e C_{2} O_{4}$ solution is mixed in some water to make a solution containing 900g of water. Which of the following is/are true for this solution?
$(K_{f} for water = 1.8 / Kkg / mol)$

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The RLVP of the solution is approximately $2 \times 10^{- 3}$

Osmotic pressure of the initial $F e C_{2} O_{4}$ solution is 49.2 atm at 300 K

Osmotic pressure of the initial $K M n O_{4}$ solution (without acid) is 19.68 atm at 300 K

Freezing point of the solution is 0.2 K

answer is A, B, C.

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Detailed Solution

$F e C_{2} O_{4} + K M n O_{4} \to_{}^{} M n^{2 +} + F e^{3 +} + C O_{2}$

$\Rightarrow 10 \times 3 \times M = 30 \times 0.4 \times 5$

$M = 2$

$\Rightarrow π = 2 \times 0.82 \times 300 \times 2 = 49.2 a t m \times 2 = 98.4 a t m$

$π_{K M n O_{4}} = 0.4 \times 0.082 \times 300 \times 2 = 19.68 a t m$

In the new $F e C_{2} O_{4}$ solution

$n_{F e C_{2} O_{4}} = \frac{25}{1000} \times 2 = \frac{1}{20}$

$R L V P = \frac{\frac{1}{20} \times 2}{\frac{1}{20} \times 2 + \frac{900}{18}} = \frac{\frac{1}{10}}{\frac{1}{10} + 50} = \frac{1}{500}$

and

$Δ T_{1} = 1.8 \times \frac{\frac{1}{20}}{0.9} \times 2 = 2 \times 1.8 \times \frac{1}{18} = 0.2 K$

$T_{f} = - 0.2 K$

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