Question

The number of kilograms of wet $NaOH$ containing $12\%$ water required to prepare $60$ litres of $0.5N$ solution is (as the nearest integer) :

Answer 1

Meq. of $NaOH$ required $=0.5\times 60000=30000$

$\therefore \frac{W_{NaOH}}{40}\times 1000=30000$

$W_{NaOH}$ required $=1200$ g

$\because 88$ g pure $NaOH$ is needed, then mass of wet sample $=100$ g.
$\therefore 1200$ g pure $NaOH$ is needed, then mass of wet sample$=\frac{(100\times 1200)}{88}=1364$ g.

Mass of $NaOH$ required $=1.364$ kg

So, the nearest integer is $1$.