Three different solutions are prepared by adding 'w' gm of 'A' into 1 kg of water, 'w' gm of 'B' into another 1 kg of water and 'w' gm of 'C' in another 1 kg of water (A, B, C are non electrolytic). Dry air is passed over these solutions in sequence $(A \to B \to C)$ . The loss in weight of solution A was found to be 2 gm while solution B gained 0.5 gm and solution C lost 1 gm. Then the relation between molar masses of A, B and C is/are :

M_{A} : M_{B} : M_{C} = 4 : 3 : 5

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M_{A} : M_{B} : M_{C} = \frac{1}{4} : \frac{1}{3} : \frac{1}{5}

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M_{C} > M_{A} > M_{B}

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M_{B} > M_{A} > M_{C}

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Solution

The correct option is B $M_{A} : M_{B} : M_{C} = \frac{1}{4} : \frac{1}{3} : \frac{1}{5}$

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The correct option is B MA:MB:MC=14:13:15

The correct option is B $M_{A} : M_{B} : M_{C} = \frac{1}{4} : \frac{1}{3} : \frac{1}{5}$