A beaker containing `20 g` sugar in `100 g` water and another containing `10g` sugar in `100 g` water are placed under a bell-jar and allowed to stand until equilibrium is reached. How much water will be transferred from one beaker to other?

Explanation

At equilibrium, both solutions will have the same vapour pressure which will be so when they have the same molar concentration. Thus, some water from dilute solution will be transferred to concentrated solution. Suppose w g of water is transferred from beakder 1 containing 10 g sugar/100 g of water to the beaker 2 containing 20 g sugar/100 g of water. Taking the density of both the solutions to be nearly same `(="d g mL"^(-1))`
Molar concentration of sugar in beaker 1
`=(10)/(342)xx(1)/(((100-w))/(d))xx1000=(10)/(342)xx(1000d)/(1000-w)"mol L"^(-1)`
Molar concentration of sugar in beaker 2
`=(20)/(342)xx(1000d)/(1000+w)"mol L"^(-1)`
`therefore" "(10)/(342)xx(1000d)/(100-w)=(20)/(342)xx(1000d)/(1000+w)`
`"or "(10)/(100-w)=(20)/(100+w)" or "(1)/(100-w)=(2)/(100+w)" or "100+w=200-2w`
`"or "3w=100" or "2=33.3g`