Toluene in the vapour phase is in equilibrium with a solution of benzene and toluene having mole fraction of toluene $0.50$. If vapour pressure of pure benzene is $119 \ torr$ and that of toluene is $37.0 \ torr$ at the same temperature,mole fraction of toluene in vapour phase will be:
- A$0.462$
- B$0.237$
- C$0.506$
- D$0.325$
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$X$ is a non-volatile solute and $Y$ is a volatile solvent. The following vapour pressures are observed by dissolving $X$ in $Y$ at different concentrations:
| $X / \text{mol L}^{-1}$ | $Y / \text{mm of Hg}$ |
| :--- | :--- |
| $0.10$ | $p_1$ |
| $0.25$ | $p_2$ |
| $0.01$ | $p_3$ |
The correct order of vapour pressures is:
| $X / \text{mol L}^{-1}$ | $Y / \text{mm of Hg}$ |
| :--- | :--- |
| $0.10$ | $p_1$ |
| $0.25$ | $p_2$ |
| $0.01$ | $p_3$ |
The correct order of vapour pressures is:
The vapour pressure of pure benzene at a certain temperature is $200 \ mm \ Hg$. At the same temperature,the vapour pressure of a solution containing $2 \ g$ of a non-volatile,non-electrolyte solid in $78 \ g$ of benzene is $195 \ mm \ Hg$. What is the molecular weight of the solid?
Medium
View SolutionWhen a solute is added to a solvent having a vapor pressure of $0.80 \ atm$,the vapor pressure decreases to $0.60 \ atm$. The mole fraction of the solute is .........
Easy
View SolutionThe vapour pressures of two liquids $A$ and $B$ in their pure states are in the ratio of $1:2$. $A$ binary solution of $A$ and $B$ contains $A$ and $B$ in the mole proportion of $1:2$. The mole fraction of $A$ in the vapour phase of the solution will be
The vapour pressure of acetone at $20\,^{\circ}C$ is $185\,torr.$ When $1.2\,g$ of a non-volatile substance was dissolved in $100\,g$ of acetone at $20\,^{\circ}C,$ its vapour pressure was $183\,torr.$ The molar mass $(g\,mol^{-1})$ of the substance is:
DifficultJEE MAIN 2015
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