$18 \ g$ glucose $(C_6H_{12}O_6)$ is added to $178.2 \ g$ water. The vapour pressure of this aqueous solution at $100 \ ^{\circ}C$ (in $torr$) is:

$224 \, mL$ of $SO_{2(g)}$ at $273 \, K$ and $1 \, atm$ is passed through $100 \, mL$ of $0.1 \, M \, NaOH$ solution. The non-volatile solute produced is dissolved in $36 \, g$ of water. The lowering of vapour pressure of the solution (assuming the solution is dilute) $(P^0_{H_2O} = 24 \, mm \, Hg)$ is $x \times 10^{-2} \, mm \, Hg$. The value of $x$ is ...... . (Integer answer)

$160 \ g$ of non-volatile solute '$A$' is dissolved in $54 \ mL$ of water at $373 \ K$. What is the vapour pressure of the aqueous solution of '$A$' (in $Torr$)? (Given: molecular weight of '$A$' = $160 \ g \ mol^{-1}$)

What is the vapour pressure of a solution containing $1.8 \ g$ of glucose in $16.2 \ g$ of water,if the vapour pressure of pure water is $32 \ mm \ Hg$ (in $mm \ Hg$)?

$X$ is a non-volatile solute and $Y$ is a volatile solvent. The following vapour pressures are observed by dissolving $X$ in $Y$ at different concentrations:
| $X / \text{mol L}^{-1}$ | $Y / \text{mm of Hg}$ |
| :--- | :--- |
| $0.10$ | $p_1$ |
| $0.25$ | $p_2$ |
| $0.01$ | $p_3$ |
The correct order of vapour pressures is: