A solution is made by mixing 1 mole of volat | vedclass

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

Question

(D) According to Raoult's Law,the total vapour pressure of the solution is given by $P_{S} = P_{A}^{o} \cdot X_{A} + P_{B}^{o} \cdot X_{B}$.Given: $n_

Vedclass · Accepted Answer

$600 \ mm \ Hg, A$

Vedclass · Answer

(D) According to Raoult's Law,the total vapour pressure of the solution is given by $P_{S} = P_{A}^{o} \cdot X_{A} + P_{B}^{o} \cdot X_{B}$.Given: $n_{A} = 1 \ mole$,$n_{B} = 3 \ moles$,$P_{A}^{o} = 200 \ mm \ Hg$,$P_{S} = 500 \ mm \ Hg$.Mole fractions are $X_{A} = \frac{1}{1+3} = 0.25$ and $X_{B} = \frac{3}{1+3} = 0.75$.Substituting the values: $500 = (200 	imes 0.25) + (P_{B}^{o} 	imes 0.75)$.$500 = 50 + 0.75 \cdot P_{B}^{o}$.$450 = 0.75 \cdot P_{B}^{o} \Rightarrow P_{B}^{o} = \frac{450}{0.75} = 600 \ mm \ Hg$.Since $P_{A}^{o} (200 \ mm \ Hg) < P_{B}^{o} (600 \ mm \ Hg)$,component $A$ is the least volatile.

Similar Questions

Which of the following changes will cause an increase in the vapour pressure of a $1 \ m$ aqueous $KI$ solution at the same temperature?

$38.4 \ g$ of unknown substance (molar mass $384 \ g \ mol^{-1}$) and $116 \ g$ of acetone is used to prepare a solution at $313 \ K$. If vapour pressure of pure acetone (molar mass $58 \ g \ mol^{-1}$) is $0.842 \ atm$,what is the vapour pressure of the solution (in $atm$)?

What is the relation between the vapour pressure of the solution,the vapour pressure of the pure solvent,and its mole fraction in the solution?

Vedclass Test Series

Exam Paper Generator

Online Exam Module