Solveeit Logo

Question

Question: An ideal solution contains two volatile liquids \(A\) \((p = \) \(100\) torr) and \(B\) \((p = 200\)...

An ideal solution contains two volatile liquids AA (p=(p = 100100 torr) and BB (p=200(p = 200 torr)torr). If the mixture contains 11 mole of AA and 44 moles of BB then total vapour pressure of the distillate is:
A.A. 150150
B.B. 180180
C.188.88C.188.88
D.198.88D.198.88

Explanation

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 AA (nA)({n_A}) == 11 molemole.
No of Moles of BB (nB)({n_B}) == 44 molesmoles.
As we know that,
Mole fraction=No. of molesTotal moles\text{Mole fraction} = \dfrac{\text{No. of moles}}{\text{Total moles}}
Therefore
XA{X_A} == 11+4\dfrac{1}{{1 + 4}} == 0.20.2.
XB{X_B} == 41+4\dfrac{4}{{1 + 4}} == 0.80.8 .
Again, as we know that,
PT{P_T} == PA{P_A} ++ PB{P_B}
\Rightarrow PT=poa×Xa+pob×Xb{P_T} = {p^o}_a \times {X_a} + {p^o}_b \times {X_b}
pA{p^ \circ }A == 100100 torrtorr
pB{p^ \circ }B == 200200 torrtorr
PT{P_T} == (0.2×100)+(0.8×200)(0.2 \times 100) + (0.8 \times 200)
\Rightarrow PT{P_T} == 20+16020 + 160 =180 = 180torr.
So the correct Answer is B.180B.180.
Hence the total vapour pressure of the distillate is 180180 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.