A student has $100 \mathrm{ml}$  of  $0.1 \mathrm{M} \mathrm{KCl}$ solution. To make it $0.2 \mathrm{M}$ , he has to

We know that

Molarity $=\frac{\mathrm{No}. \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{Volume} \mathrm{of} \mathrm{solution}(\mathrm{in} \mathrm{litres})}$

Molarity $=\frac{\mathrm{n}}{0.1 \mathrm{L}}=0.2\mathrm{M}$ $(1\mathrm{l}=1000 \mathrm{ml})$

Hence, $\mathrm{n}=0.02$

The number of moles should be increased from $0.01$ to $0.02$ by the addition of 0.01 moles of $\mathrm{KCl}$.

$$\begin{gathered} \text{Also } \\ 100 \mathrm{ml}\times 0.1\mathrm{M}=\mathrm{xml}\times 0.2\mathrm{M} \\ \mathrm{x}=\frac{100 \mathrm{ml}\times 0.1\mathrm{M}}{0.2\mathrm{M}}=50 \mathrm{ml} \end{gathered}$$

Hence to make the required solution, the total volume of the solution should be reduced to $50 \mathrm{ml}$.

This can be done by evaporating $50 \mathrm{ml}$ of solution.

Hence both $\mathrm{a}$ and $\mathrm{b}$ are correct which is option $\left(\mathrm{c}\right)$.