Non-volatile non electrolyte solute X(M.W=100) in solvent ‘S’ behaves as ideal solution. Non-volatile solute Y(M.W=60) undergo 100% association in solvent S and this solution behaves ideally,

From the above vapour pressure vs temperature graphs correct statements(s) among the following is/are

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Osmotic pressure of 1M Solution of Y in S is 2RT.

In the solvent S, Y undergo dimerisation.

When 1060 g of solution (mentioned in graph-II) of Y in S is cooled to $18 . 5^{ο} C$ , 560 g of liquid solution left and 500 g of solid S separated.

When 1100 g of solution (mentioned in graph-I) of X in S is heated to $114^{ο} C$ at the time of stopping boiling 350 g of liquid solution left behind.

answer is A, B, C.

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Detailed Solution

From the first graph Kb of S is 1K.kg/mol, Kf of S is 1.5K.kg/mol.
A) For 1m Y in S $Δ T_{f} = 0.75 K, Δ T_{b} = 0.5 K$ . Thus Y undergoing 100% dimerisation
B) $Δ T_{b} = K_{b} . m$
$4 = 1 \times \frac{100}{100} \times \frac{1000}{W} \Rightarrow$ wt of liquid S left = 250 g.
Wt. of liquid solution left = 250 + 100 = 350 g.
C) $Δ T_{f} = K_{f} . m . i$
$1.5 = 1.5 \times \frac{60}{60} \times \frac{1000}{W} \times 0.5 \Rightarrow$ Wt. of liquid S left = 500 g
$∴$ Weight of liquid solution left = 500 + 60 = 560 g
D) $π = C R T i = 1 \times R T \times 0.5 = 0.5 R T$