The vapour pressure of solvent decreases by 5.4 torr when a non volatile solute is added.In this silution ,mole fraction of solute is 0.2. What would be mole fraction of the solvent if decreses in vapour pressure is 16.2 torr?
Answers
Answered by
32
According to Raoult's law:
The relative lowering of vapour pressure is equal to the mole fraction of solute.
So, (p - p₀)/p = n/ (n + N)
or Δp / p = n/ (n + N)
So, 5.4 / p = 0.2
p = 27 torr
For other solution of same solvent,
Δp = 16.2 torr
16.2 / 27 = n/ (n + N)
0.6 = n/ (n + N)
Taking n + N = 1, to calculate mole fraction,
0.6 is the mole fraction of solvent. And 0.4 is the mole fraction of solute.
The relative lowering of vapour pressure is equal to the mole fraction of solute.
So, (p - p₀)/p = n/ (n + N)
or Δp / p = n/ (n + N)
So, 5.4 / p = 0.2
p = 27 torr
For other solution of same solvent,
Δp = 16.2 torr
16.2 / 27 = n/ (n + N)
0.6 = n/ (n + N)
Taking n + N = 1, to calculate mole fraction,
0.6 is the mole fraction of solvent. And 0.4 is the mole fraction of solute.
Answered by
1
Answer:
answer is 0.4
hope it hepls
Similar questions