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  • Three solutions are prepared by adding ‘w’ g of ‘A’ into 1 kg of water, ‘w’ g of ‘B’ into another 1 kg of water and ‘w’ g 

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

Option: 1

M_A : M_B : M_C = 4:3:5


Option: 2

M_A : M_B : M_C = \frac{1}{4}:\frac{1}{3}:\frac{1}{5}


Option: 3

M_C > M_A > M_B


Option: 4

M_B > M_A > M_C


Answers (1)

best_answer

The loss in weight from A is due to dry air and proportional to vapour pressure above that solution. The loss or gain in mass from solution B or C will depend upon the Pressure above that solution and the pressure inside the solution from which it is coming into the particular solution.

So,

P_{A}\propto 2gm

(P_A-P_{B})\propto 0.5gm

(P_C-P_B)\propto 2.5gm

Hence, the order of Pressure: PC > PA> PB

So, maximum vapour pressure is above solution C hence, it is having minimum lowering and hence the minimum mole fraction (hence minimum number of moles) of solute So, it has maximum molar mass of substance. 

Posted by

himanshu.meshram

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