The mass of a non-volatile solute of molar mass $40 \ g \ mol^{-1}$ that should be dissolved in $114 \ g$ of octane to lower its vapour pressure by $20 \%$ is (in $g$)

At a certain temperature,when some amount of glucose is dissolved in water,it is observed that the lowering in vapour pressure is $0.6 \ mm \ Hg$ for this solution. What will be the vapour pressure of the same glucose solution if its molality is $(\frac{1}{18}) \ mol \ kg^{-1}$ (in $mm \ Hg$)?

$A$ graph of vapour pressure and temperature for three different liquids $X$,$Y$ and $Z$ is shown below.
The following inferences are made:
$(A)$ $X$ has higher intermolecular interactions compared to $Y$.
$(B)$ $X$ has lower intermolecular interactions compared to $Y$.
$(C)$ $Z$ has lower intermolecular interactions compared to $Y$.
The correct inference$(s)$ is/are:

Which one of the following is the incorrect form of Raoult's law?

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 $180 \ g \ mol^{-1}$

Calculate the relative lowering of vapour pressure if the vapour pressure of the pure solvent and the vapour pressure of the solution at $25^{\circ} C$ are $32 \ mm \ Hg$ and $30 \ mm \ Hg$ respectively.