Vapour pressure of a pure solvent at $90\,^{\circ}C$ is $1020\, torr$. $A$ solution of $0.1\, m$ concentration has a vapour pressure of $1000\, torr$. What is the molecular weight of the solvent?

The vapour pressure of water at $23^{\circ} C$ is $19.8 \ mm$. If $0.1 \ mole$ of glucose is dissolved in $178.2 \ g$ of water, what is the vapour pressure (in $mm$) of the resultant solution?

At $50^{\circ} C$, the vapour pressure of pure benzene is $268 \ torr$. The number of moles of non-volatile solute per mole of benzene required to prepare a solution having a vapour pressure of $167 \ torr$ at the same temperature is (molar mass of benzene $= 78 \ g \ mol^{-1}$)

The vapour pressure of water at room temperature is lowered by $5\%$ by dissolving a solute in it. Then the approximate molality of the solution is

When approximately $100 \ mL$ of acetone is added to $1 \ L$ of water,the vapor pressure of water in the solution will be............

$A$ 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.