The vapour pressure of a solvent decreased by 10 mm ofmercury when a non‐volatile solute was added to the solvent. The mole fraction ofthe solute in the solution is 0.2. What should be the mole fract…

Question

The vapour pressure of a solvent decreased by 10 mm ofmercury when a non‐volatile solute was added to the solvent. The mole fraction ofthe solute in the solution is 0.2. What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be 20 mm of mercury?

Accepted Answer

Correct answer: B. (b) x_1 (mole fraction of solute) =0.2 From Raoult’s law, P^o−P_sP^o=x_1= 1 0P^o=0.2=P^o=5 0 mm Again, when P^o−P_s=2 0 mm, then P^o−P_s 20 =mole fr action of solute =—=0.4P^o 50 =mole fraction of solvent =1−0.4=0.6

$The vapour pressure of a solvent decreased by 10 mm ofmercury when a non‐volatile solute was added to the solvent. The mole fraction ofthe solute in the solution is 0.2. What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be 20 mm of mercury?$

Solution

More Solutions PYQs