The vapor pressure of water at $23\,^{\circ}C$ is $19.8\,mm\,Hg$. What will be the vapor pressure (in $mm\,Hg$) of the resulting solution when $0.1\,mol$ of glucose is dissolved in $178.2\,g$ of water?

$X$ 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:

What weight of glucose must be dissolved in $100 \ g$ of water to lower the vapour pressure by $0.20 \ mm \ Hg$ (in $g$)? (Assume dilute solution is being formed) Given: Vapour pressure of pure water is $54.2 \ mm \ Hg$ at room temperature. Molar mass of glucose is $180 \ g \ mol^{-1}$

What is the molar mass of a solute when $2.3 \ g$ of a non-volatile solute is dissolved in $46 \ g$ of benzene at $30^{\circ} C$? (Relative lowering of vapour pressure is $0.06$ and the molar mass of benzene is $78 \ g \ mol^{-1}$)

At $25\,^{\circ}C$,the vapour pressure of pure liquid $A$ $(mol. wt. = 40)$ is $100\, torr$,while that of pure liquid $B$ is $40\, torr$ $(mol. wt. = 80)$. The vapour pressure at $25\,^{\circ}C$ of a solution containing $20\, g$ of each $A$ and $B$ is .......... $torr$.

$A$ solution containing $30 \, g$ of non-volatile solute in exactly $90 \, g$ water has a vapour pressure of $21.85 \, mm \, Hg$ at $25 \, ^oC$. Further $18 \, g$ of water is then added to the solution. The resulting solution has a vapour pressure of $22.15 \, mm \, Hg$ at $25 \, ^oC$. Calculate the molecular weight of the solute.