A solution at 20,^oC is composed of 1.5,mol o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $A$ be benzene and $B$ be toluene. Given: $P^o_A = 74.7\,torr$,$P^o_B = 22.3\,torr$,$n_A = 1.5\,mol$,$n_B = 3.5\,mol$. Total moles $n_{total}

Vedclass · Accepted Answer

$38.0\,torr$ and $0.589$

Vedclass · Answer

(B) Let $A$ be benzene and $B$ be toluene. Given: $P^o_A = 74.7\,torr$,$P^o_B = 22.3\,torr$,$n_A = 1.5\,mol$,$n_B = 3.5\,mol$. Total moles $n_{total} = 1.5 + 3.5 = 5.0\,mol$. Mole fraction of benzene in liquid phase,$x_A = \frac{1.5}{5.0} = 0.3$. Mole fraction of toluene in liquid phase,$x_B = \frac{3.5}{5.0} = 0.7$. Total vapour pressure of the solution,$P_{total} = P^o_A x_A + P^o_B x_B = (74.7 	imes 0.3) + (22.3 	imes 0.7) = 22.41 + 15.61 = 38.02\,torr \approx 38.0\,torr$. Mole fraction of benzene in vapour phase $(y_A)$ is given by $y_A = \frac{P_A}{P_{total}} = \frac{P^o_A x_A}{P_{total}} = \frac{22.41}{38.02} \approx 0.589$.

Similar Questions

In a mixture of camphor in nitrogen gas,the physical states of the solute and the solvent are,respectively:

At $80\,^oC$,the vapour pressure of pure liquid $A$ is $520\, mm\,Hg$ and that of pure liquid $B$ is $1000\, mm\,Hg$. If a mixture of $A$ and $B$ boils at $80\,^oC$ and $1\, atm$ pressure,the amount of $A$ in the mixture is ........... $mol\,\%$.

An aqueous solution of urea:

An aqueous solution of a solute $AB$ has a normal boiling point of $101.08\,^{\circ}C$ and a normal freezing point of $-1.80\,^{\circ}C$. Hence,$AB$:
Given: $AB$ is $100\%$ ionised at the boiling point of the solution,$(K_b / K_f)_{\text{water}} = 0.3$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module