Question

How many kg (nearest integer value) of wet $NaOH$ containing $12\%$ water is required to prepare $60$ L of $0.50M$ $NaOH$ solution? If the answer is $x$ then the nearest integral value of $1000x$.

Answer 1

The number of moles of $NaOH$ in $60$ L of $0.50$ M pure $NaOH$ solution $=60\times 0.50=30$ mol

This corresponds to $30\times 40=1200$ g of pure $NaOH$.

However, $NaOH$ contains $12\%$ water.

Hence, the mass of $NaOH$ is $\frac{1200\times 100}{88}=1363.64$ g wet $NaOH$.

Hence, mass of wet $NaOH$ is $1.364kg$.

$\therefore 1000x=1364$