The pH of a sample of $\mathrm{KOH}$ solution is 12.3979. The weight of solid $\mathrm{KOH}$ of $70\%$ pure required to prepare $2.5$ lit of this solution is

The $pOH$ of the solution will be:

$pOH$=$14-12.3979=1.6021$

The normality of the solution is $2.5\times {10}^{-2}$ .

The weight can be calculated as $N=\frac{w}{eq.\text{ wt}}\times 1000$.

Putting the values, we get:

 $2.5$$\times {10}^{-2}=\frac{w}{56}\times 2.5$

 $w=56$$\times 2.5\times 2.5\times {10}^{-2}=3.5g$

70% pure KOH means:

  $\frac{100}{70}=\frac{pure}{3.5}$

 $pure=$$\frac{100}{70}\times \frac{35}{10}=5g$

Therefore, the correct option is (B).