The vapour pressure of a solvent $A$ is $0.80 \, atm$. When a non-volatile substance $B$ is added to this solvent its vapour pressure drops to $0.6 \, atm$. What is the mole fraction of $B$ in the solution?

The vapour pressure of water at $20\,^{\circ}C$ is $17.54\,mm$. When $20\,g$ of a non-ionic substance is dissolved in $100\,g$ of water,the vapour pressure is lowered by $0.30\,mm$. What is the molecular weight of the substance?

$A$ 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.

When $0.25 \text{ moles}$ of a non-volatile,non-ionizable solute was dissolved in $1 \text{ mole}$ of a solvent,the vapor pressure of the solution was $x\%$ of the vapor pressure of the pure solvent. What is $x$ (in $\%$)?

$A$ 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:

If $8 \; g$ of a non-electrolyte solute is dissolved in $114 \; g$ of $n$-octane to reduce its vapour pressure to $80 \%$,the molar mass (in $g \; mol^{-1}$) of the solute is.
Given that the molar mass of $n$-octane is $114 \; g \; mol^{-1}$.