The vapour pressure of water at room temperature is $23.8 \ mm \ Hg$. The vapour pressure of an aqueous solution of sucrose with mole fraction of $0.1$ is equal to .......... $mm \ Hg$.

How many grams of a non-volatile solute (molar mass $40 \ g/mol$) must be added to $114 \ g$ of octane to reduce its vapour pressure by $20\%$ (in $g$)?

Vapour pressure of a solution containing $18 \ g$ of glucose and $178.2 \ g$ of water at $100^{\circ} C$ is (Vapour pressure of pure water at $100^{\circ} C = 760 \ torr$) (in $torr$)

Calculate the weight of non-volatile solute having molecular weight $40$,which is dissolved in $57 \ g$ octane to reduce its vapour pressure by $20\%$.

When $1 \ g$ of urea is dissolved in $50 \ g$ of water,the relative lowering of vapour pressure of the solution is equal to the relative lowering of vapour pressure of a solution formed by dissolving glucose in the same amount of water. How many grams of glucose were dissolved?

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