For first order reaction

k = ${{2.303} \over t}\log {{{{\left[ A \right]}_0}} \over {{{\left[ A \right]}_t}}}$

$\Rightarrow$ k = ${{2.303} \over 1}\log {{0.8} \over {0.2}}$ = 2.303 log 4 ....(1)

Let t₁ hour is required for changing the concentration of A from 0.9 mole to 0.675 mole of B.

Remaining mole of A = 0.9 – 0.675 = 0.225

$\therefore$ k = ${{2.303} \over {{t_1}}}\log {{0.9} \over {0.225}}$ .....(2)

From equation (1) and (2)

${{2.303} \over {{t_1}}}\log {{0.9} \over {0.225}}$ = 2.303 log 4

$\Rightarrow$ ${{t_1}}$ = 1 hr