$100{\mathrm{cm}}^3$ of 0.1 M HCl and $100c{m}^3$ of $0.1\mathrm{M}\mathrm{NaOH}$ solutions are mixed in a calorimeter. If the heat liberated is "Q" K.Cal, the heat of neutralization ( $\Delta \mathrm{H}$ ) (in $\mathrm{K}$.Cal) of $\mathrm{HCl}\left(\mathrm{aq}\right)$ and $\mathrm{NaOH}\left(\mathrm{aq}\right)$ is

We can calculate the milliequivalents of NaOH and HCl as:

$\mathrm{Milli} \mathrm{equivalents}\text{ of NaOH = 100 }\times \text{ 0.1 =10}$ 

$\mathrm{Milli} \mathrm{equivalents}\text{ of HCl = 100 }\times \text{ 0.1 =10}$ 

As a result, the amount of heat generated by neutralizing 10 milliequivalents of $\mathrm{HCl}$ with 10 milliequivalents of $\mathrm{NaOH}$ is $\mathrm{x}\mathrm{KJ}$. However, the process of neutralizing one equivalent of $\mathrm{HCl}$ with one equivalent of $\mathrm{NaOH}$ generates heat. Therefore,

$\mathrm{\Delta H}=-\frac{\mathrm{Q}}{10\times {10}^{-3}}=-100\mathrm{Q}\text{ kJ/mol}$ 

Therefore, the correct option is B.