$A$ gaseous mixture of two substances $A$ and $B$,under a total pressure of $0.8 \ atm$ is in equilibrium with an ideal liquid solution. The mole fraction of substance $A$ is $0.5$ in the vapour phase and $0.2$ in the liquid phase. The vapour pressure of pure liquid $A$ is $...... \ atm$. (Nearest integer)
- A$1$
- B$0$
- C$2$
- D$3$
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At $300 \ K$,the vapour pressures of two pure liquids,$A$ and $B$ are $100 \ mm \ Hg$ and $500 \ mm \ Hg$,respectively. If in a mixture of $A$ and $B$,the total vapour pressure is $300 \ mm \ Hg$,the mole fractions of $A$ in the liquid and in the vapour phase,respectively,are:
MediumKVPY 2011
View SolutionWhat is the vapour pressure of a solution containing $1 \ mol$ of a non-volatile solute in $36 \ g$ of water $\left(P_1^0 = 32 \ mm \ Hg\right) (in $mm \ Hg$)?$
In a solution,the mole fraction of solute is $0.2$ when the lowering in vapour pressure is $10 \text{ mm Hg}$. To achieve a lowering of vapour pressure of $20 \text{ mm Hg}$,the mole fraction of the solute in the solution should be:
The vapor pressures of two liquids $P$ and $Q$ are $80 \ torr$ and $60 \ torr$,respectively. The total vapor pressure of a solution formed by mixing $3 \ moles$ of $P$ and $2 \ moles$ of $Q$ is .......... $torr$.
Medium
View SolutionThe vapour pressure of benzene at a certain temperature is $640 \, mm$ of $Hg$. $A$ non-volatile and non-electrolyte solid weighing $2.175 \, g$ is added to $39.08 \, g$ of benzene. The vapour pressure of the solution is $600 \, mm$ of $Hg$. What is the molecular weight of the solid substance?
DifficultAIPMT 1999
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