A gas described by van der Waals equation (multiple correct answers type)

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 freezing point of the solution $\mathrm{M}$ is

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

A solution 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$=46 \mathrm{g} {\mathrm{mol}}^{-1}$ 

Water is added to the solution M such that the mole fraction of water in solution becomes 0.9. What is the boiling point of this solution?

[Hint: Solute is ethanol and solvent is water]

Electrolysis of dilute aqueous $\mathrm{NaCl}$ solution was carried out by passing a current of$10$ $\mathrm{milliamperes}$. The time required to liberate $0.01$ $\mathrm{moles}$ of ${\mathrm{H}}_2$ gas at the cathode is $\left(1 \mathrm{F}=96500 \mathrm{C} {\mathrm{mol}}^{-1}\right)$.

How many effective numbers of atoms (spheres) are contained within the hexagonal prism shown below in the $\mathrm{HCP}$ space lattice?

The cell content$\left(Z\right)$ of the unit cell of a $\mathrm{HCP}$ space lattice is

The volume of the hexagonal prism is (radius of every sphere is $r)$