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Q.

A set of solutions is prepared using 180g of water as a solvent and 10g of different non-volatile solutes A, B and C. The relative lowering of vapour pressure in the presence of these solutes are in the order [Given molar mass of A = 100g $m o l^{- 1}$ ; B = 200 g $m o l^{- 1}$ ; C = 10,000g $m o l^{- 1}$ ]

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a

B > C > A

b

C > B > A

c

A > C > B

d

A > B > C

answer is D.

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Detailed Solution

$Relative lowering of V.P. = \frac{Δ P}{P^{0}} = x_{solute} {(\frac{ΔP}{P^{0}})}_{A} = \frac{\frac{10}{100}}{\frac{10}{100} + \frac{180}{18}} : {(\frac{ΔP}{P^{0}})}_{B} = \frac{\frac{10}{200}}{\frac{10}{200} + \frac{180}{18}} {(\frac{ΔP}{P^{0}})}_{C} = \frac{\frac{10}{10, 000}}{\frac{10}{10, 000} + \frac{180}{18}} : {(\frac{ΔP}{P^{0}})}_{A} > {(\frac{ΔP}{P^{0}})}_{B} > {(\frac{ΔP}{P^{0}})}_{C}$

Hence, the correct option is: (D) A > B > C

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A set of solutions is prepared using 180g of water as a solvent and 10g of different non-volatile solutes A, B and C. The relative lowering of vapour pressure in the presence of these solutes are in the order [Given molar mass of A = 100g mol-1 ; B = 200 g mol-1 ; C = 10,000g mol-1]