$$\therefore$$ $${K_C} = {\left( {{{1 + x} \over {1 - x}}} \right)^2}$$
$$100 = {\left( {{{1 + x} \over {1 - x}}} \right)^2}$$
$${{1 + x} \over {1 - x}} = 10$$
$$x = {9 \over {11}}$$
Moles of D = 1 + x
$$ = 1 + {9 \over {11}} = {{20} \over {11}}$$
$$ = 1.818 = 181.8 \times {10^{ - 2}} = 181.8 \times {10^{ - 2}}$$
$$ \cong 182 \times {10^{ - 2}}$$ M
