$A$ solution was prepared by dissolving $0.1 \ mol$ of a nonvolatile solute in $0.9 \ mol$ of water. What is the relative lowering of vapour pressure of the solution?

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 $78 \ g \ mol^{-1}$)

When $25 \ g$ of a non-volatile solute is dissolved in $100 \ g$ of water, the vapour pressure is lowered by $2.25 \times 10^{-1} \ mm$. If the vapour pressure of water at $20^{\circ}C$ is $17.5 \ mm$, what is the molecular weight of the solute?

$A$ 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:

The vapour pressures of two liquids $A$ and $B$ in their pure states are in the ratio of $1:2$. $A$ binary solution of $A$ and $B$ contains $A$ and $B$ in the mole proportion of $1:2$. The mole fraction of $A$ in the vapour phase of the solution will be

The vapor pressure of a pure liquid solvent $(X)$ decreases from $0.80 \ atm$ to $0.60 \ atm$ upon the addition of a non-volatile solute $(Y)$. What is the mole fraction of $(Y)$ in the solution?