Question
Question: An ideal solution contains two volatile liquids \(A\) \((p = \) \(100\) torr) and \(B\) \((p = 200\)...
An ideal solution contains two volatile liquids A (p= 100 torr) and B (p=200 torr). If the mixture contains 1 mole of A and 4 moles of B then total vapour pressure of the distillate is:
A. 150
B. 180
C.188.88
D.198.88
Solution
The vapour pressure of a solvent above a solution equals the vapour pressure of a pure solvent at the same temperature scaled by the mole fraction of the solvent present, according to Raoult's law. As a result, the mole fraction of nonvolatile solute in a dilute solution lowers the vapour pressure compared to the mole fraction of solute in the solution.
Complete Answer:
No of Moles of A (nA) = 1 mole.
No of Moles of B (nB) = 4 moles.
As we know that,
Mole fraction=Total molesNo. of moles
Therefore
XA = 1+41 = 0.2.
XB = 1+44 = 0.8 .
Again, as we know that,
PT = PA + PB
⇒ PT=poa×Xa+pob×Xb
p∘A = 100 torr
p∘B = 200 torr
PT = (0.2×100)+(0.8×200)
⇒ PT = 20+160 =180torr.
So the correct Answer is B.180.
Hence the total vapour pressure of the distillate is 180 torr.
Additional Information:
It's worth noting that Raoult's law isn't always followed in practise across the entire concentration range. Ideal solutions are those that obey Raoult's law across the whole concentration range, and most solutions deviate from the ideal behaviour in either a positive or negative way. For a volatile liquid, it offers us the relationship between mole fraction and vapour pressure.
Note:
The partial pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction in the mixture, according to Raoult's law. The mole fraction of nonvolatile solute in a dilute solution lowers the vapour pressure compared to the mole fraction of solute in the solution.