$1 {M}\quad 1 {M} \quad 1 {M} \quad1 {M}$

First check direction of reversible reaction.

Since ${Q}_{{c}}=\frac{[{C}][{D}]}{[{A}][{B}]}=1<{K}_{{eq} .} \Rightarrow$ reaction will move in forward direction to attain equilibrium state.

$\Rightarrow {A}+{B} \rightleftharpoons {C}+{D}: {K}_{{eq}}=100$

$to \quad 1\quad 1\quad\quad 1\quad 1$

${t}_{\text {eq. }} \, 1-{x} \, 1-{x} \, 1+{x} \, 1+{x}$

Now $: {K}_{{eq}}=100=\frac{(1+{x})(1+{x})}{(1-{x})(1-{x})}$

$\Rightarrow 100=\left(\frac{1+x}{1-x}\right)^{2}$

$(i)$ $10=\left(\frac{1+{x}}{1-{x}}\right)$

$\Rightarrow 10-10 {x}=1+{x}$

$\Rightarrow 11 {x}=9$

$\Rightarrow {x}=\frac{9}{11}$

$(ii)$ $-10=\frac{1+{x}}{1-{x}}$

$\Rightarrow-10+10 {x}=1+{x}$

$\Rightarrow-9 x=-11$

$\Rightarrow x=\frac{11}{9}$

$\rightarrow$ $'x'$ cannot be more than one, therefore not valid. therefore equation concretion of $(D)=1+x$

$=1+\frac{9}{11}=\frac{20}{11}$

$=1.8181=181.81 \times 10^{-2}$

$\simeq 182 \times 10^{-2}$