Question

How many kilograms of wet NaOH containing $12\%$ water are required to prepare 60 litres of a solution of $0.50N$ the solution?

Answer 1

The normality of the solution is defined as the gram equivalent weight of the solute per litre of the solution. Wet NaOH contains $12\%$ water and $88\%$ NaOH solution. It is also known that $1L = 1000ml$

Complete step by step answer:
It is known that the normality of the solution is defined as the gram equivalent weight of the solute per litre of the solution. The normality of the solution can be calculated as,
$Normality = \dfrac{{x(g) \times 1L}}{{Eq.wt \times y(L)}}$
Where,
x-mass of solute in gram
eq. wt-equivalent weight of the solute
y-volume of a solution in a litre
It is given that the normality of a solution is $0.50N$. Thus, the mass of NaOH can be calculated as,
$0.50 = \dfrac{{x(g) \times 1L}}{{40 \times 6(L)}}$
Equivalent weight of NaOH=$40g.eq$
$\Rightarrow 0.50 \times 40 \times 6 = x(g)$
Thus, $x = 1200g$
Therefore, $1200g$ NaOH is required to prepare $0.50N$ in a 60 litres solution.
But in the question, it is given that NaOH contains $12\%$ water and $88\%$ NaOH solution. Thus, the accurate mass of NaOH is yet to be calculated.
Thus, the accurate mass of NaOH can be calculated as,
$Mass = \dfrac{{1200g \times 100\% }}{{88\% }}$
$\Rightarrow Mass = 1363.64g$
It is known that 1 kilogram is equal to 1000 grams.
Thus, the accurate mass of NaOH in kilogram can be written as,
$\Rightarrow Mass = 1.36Kg$

Therefore, $1.36Kg$ of wet NaOH containing $12\%$ water are required to prepare 60 litres of solution of $0.50N$ solution.

Note: The important terms used to calculate the concentration of the solution are Normality, Molality, Molarity, Mole fraction and Parts per million. Sodium hydroxide, NaOH is also known as caustic soda. Sodium hydroxide is a strong base. It is mainly used in the formation of sodium salts in the chemical reaction. It is mainly used for cleaning purposes, water treatment, and also for soap/paper production.