What is the vapour pressure of a solution containing $0.1 \ mol$ of non-volatile solute dissolved in $16.2 \ g$ of water (in $mm \ Hg$)? $(P_1^{\circ} = 32 \ mm \ Hg)$

The number of moles of chlorobenzene and bromobenzene are $0.1$ and $0.2$ respectively. The vapour pressure of pure chlorobenzene and pure bromobenzene are $0.350 \ bar$ and $0.500 \ bar$ respectively. Calculate the total vapour pressure of the solution prepared by mixing these two components.

Which of the following substances,when dissolved in water,causes the maximum decrease in the vapor pressure of water?

The vapour pressure of pure water is $23 \text{ mmHg}$. The vapour pressure of an aqueous solution, which contains $10$ mass per cent of solute '$A$' having molecular weight $50$ is (in $\text{ atm}$)

In a solution,the mole fraction of solute is $0.2$ when the lowering in vapour pressure is $10 \text{ mm Hg}$. To achieve a lowering of vapour pressure of $20 \text{ mm Hg}$,the mole fraction of the solute in the solution should be:

Calculate the amount of solute dissolved in $612 \ g$ of water at $30^{\circ} C$ if molar mass of solute is $342 \ g \ mol^{-1}$ (Relative vapour pressure lowering is $0.025$ and molar mass of water $18 \ g \ mol^{-1}$ ). (in $g$)