Chemistry, asked by udeet, 1 year ago

Two liquids A and B on mixing form an ideal solution. Their vapour pressure in the pure state are 200 and 100 mm respectively. What will be mole fraction of B in the vapour phase in equilibrium with an equimolar solution of the two ? ​

Answers

Answered by ash123smart
1

Answer:

The mole fraction of B in vapor state is 0.33.

Explanation:

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

Similar questions