The density of $20\%\left[\frac{\mathrm{w}}{\mathrm{w}}\right]$ aqueous $\mathrm{NaOH}$ solution is $1.20 {\mathrm{gmL}}^{-1}$. What is the mole fraction of water? (Molar mass of $\mathrm{NaOH}=$$\left.40 {\mathrm{gmol}}^{-1}\right)$

 We know that $20\%\left[\frac{\mathrm{w}}{\mathrm{w}}\right]$ means $20 \mathrm{g}$ of $\mathrm{NaOH}$ in $80 \mathrm{g}$ of ${\mathrm{H}}_2\mathrm{O}$.

Density of $\mathrm{NaOH}$$=1.20 \mathrm{g}/\mathrm{ml}$

We know that density$=\frac{\mathrm{Mass}}{\mathrm{Volume}}$

Therefore, Volume$=\frac{\mathrm{Mass}}{\mathrm{Density}}$

 Volume$=\frac{100 }{1.20}\mathrm{ml}$ 

We know that

No. of moles$=\frac{\mathrm{Given} \mathrm{Mass}}{\mathrm{Molar} \mathrm{Mass}}$

Then,

moles of $\mathrm{NaOH}=\frac{20}{40}=\frac{1}{2}=0.5$

moles of water $=\frac{80}{18}=4.4$

Now, Mole fraction$=\frac{\mathrm{Moles} \mathrm{of} \mathrm{solute}\left(\mathrm{NaOH}\right)}{\mathrm{Total} \mathrm{moles} \mathrm{of} \mathrm{solution}}$

therefore, mole fraction of $\left(\mathrm{NaOH}\right)=\frac{0.5}{0.5+4.4}=\frac{0.5}{4.9}=0.102$

Then, Mole fraction of water $=(1-0.102)=0.898=0.917\left(\mathrm{approx}\right)$

Hence, option(b) is correct.