What is the molarity of ${\mathrm{K}}^{+}$ ion in aqueous solution contains $17.4 \mathrm{ppm} \left(\mathrm{w}\left(\mathrm{mg}\right)/\mathrm{V}\left(\mathrm{L}\right)\right)$ of ${\mathrm{K}}_2{\mathrm{SO}}_4$? 

(Molecular weight of ${\mathrm{K}}_2{\mathrm{SO}}_4$ is $174$ $\mathrm{g} {\mathrm{mol}}^{-1}$)

A solution $\mathrm{M}$ is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is $0.9$. 

Given,

$\begin{array}{l}{\mathrm{K}}_{\mathrm{f}} \left(\mathrm{water}\right)=1.86 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}\\ {\mathrm{K}}_{\mathrm{f}} \left(\mathrm{ethanol}\right)=2.0 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}\\ {\mathrm{K}}_{\mathrm{b}} \left(\mathrm{water}\right)=0.52 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}\\ {\mathrm{K}}_{\mathrm{b}} \left(\mathrm{ethanol}\right)=1.2 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}\end{array}$

Standard freezing point of water $=273 \mathrm{K}$

Standard freezing point of ethanol $=155.7 \mathrm{K}$

Standard boiling point of water $=373 \mathrm{K}$

Standard boiling point of ethanol $=351.5 \mathrm{K}$

Vapour pressure of pure water $=32.8 \mathrm{mmHg}$ 

Vapour pressure of pure ethanol $=40 \mathrm{mmHg}$ 

Molecular weight of water $=18 \mathrm{g} {\mathrm{mol}}^{-1}$

Molecular weight of ethanol $=46 \mathrm{g} {\mathrm{mol}}^{-1}$

The vapour pressure of the solution $\mathrm{M}$ is

Two liquids, $\mathrm{X}$ and $\mathrm{Y}$, form an ideal solution. At $300 \mathrm{K}$, the vapour pressure of the solution containing one mole of $\mathrm{X}$ and three moles of $\mathrm{Y}$ is $550 \mathrm{mm} \mathrm{Hg}$ . At the same temperature, if one mole of $\mathrm{Y}$ is further added to this solution, the vapour pressure of the solution increases by $10 \mathrm{mm} \mathrm{Hg}$. The vapour pressure of $\mathrm{X}$ and $\mathrm{Y}$ (in $\mathrm{mm} \mathrm{Hg}$) in their pure states will be, respectively: