x mole $\mathrm{HI}$ is taken in 10 lit vessel and heated. If the equilibrium concentration of ${\mathrm{H}}_2\left(\mathrm{g}\right)$ is $0.1 \mathrm{mol}-{\mathrm{lit}}^{-1}$, value of x and mass of I₂ at equilibrium are

The concentration of ${\mathrm{H}}_2$ will be:

$\frac{\text{Moles}}{\text{Volume}}\text{= Concentration}$

Putting the values:

$\frac{\text{Moles}}{\text{10}}\text{=0.1}$

$\text{Moles = 0.1 }\times \text{ 10 = 1}$

The reaction will be:

$\text{2HI}\rightleftharpoons {\text{H}}_{\text{2}}{\text{+I}}_{\text{2}}$

Since, the number of moles formed for iodine is 1, the concentration will be:

${\text{ [I}}_2\text{] = moles }\times \text{ molecular weight = 1 }\times \text{ 254 = 254 g}$

Therefore, the correct option is (C).