SolutionHard
Question
A beaker (A) contains 20 g sugar in 100 g water. Another beaker (B) contains 10 g of sugar in 100 g water. Both the beakers are placed under a bell jar and allowed to stand until equilibrium is reached. How much water will be transferred from one beaker to the other? Neglect the mass of water vapour present at equilibrium.
Options
A.0
B.33.33 g water will transfer from beaker−A to beaker−B.
C.33.33 g water will transfer from beaker−B to beaker−A.
D.50.00 g water will transfer from beaker−B to beaker−A.
Solution
Final vapour pressure and hence, the composition of both solutions must be same. As solution in beaker (A) has higher concentration, its vapour pressure is low. Hence, water from (B) will transfer in (A) as vapour.
$\therefore\frac{20}{200 + x} = \frac{10}{100 - x} \Rightarrow x = 33.33$
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