d
\(\quad \quad \quad \quad\quad \quad {A} \longrightarrow \quad {B}\)

\({t}=0\quad \quad \quad \quad 1 \text { mole } \quad 0\)

\({t}=100 {~min}\quad 1-{x} \quad \quad 2 {x}\)

\(\quad \quad \quad \quad \quad =0.9 {~mol} \quad =0.2 {~mol}\)

Now, \(t=\frac{t_{1 / 2}}{\ln 2} \times \frac{\left[A_{0}\right]}{\left[A_{t}\right]}\)

\(100=\frac{{t}_{1 / 2}}{\ln 2} \times \ln \frac{1}{0.9} \Rightarrow {t}_{1 / 2}=690\, {~min} . \quad\) (Taking \(\left.\ln 3=1.11\right)\)

Answer is \(700 .\) (Nearest integer).