Question

Which one of these solutions has the highest normality?

• A. $8gKOH$ per $100mL$
• B. $0.5M{H}_{2}S{O}_4$
• C. $6g$ of $NaOH$ per $100mL$
• D. $1N{H}_{3}P{O}_4$

Answer 1

Answer: (C)

$6g$ of $NaOH$ per $100mL$

Normality of $8gKOH$ per $100mL$ is:

$N=\frac{8}{56}\times \frac{1000}{100}=1.4$ N.

Normality of $0.5M{H}_{2}S{O}_4$

$N=Molarity\times basisity=0.5\times 2=1$ N.

Normality of $6g$ of $NaOH$ per $100mL$.

$N=\frac{6}{40}\times \frac{1000}{100}=1.5$ N.

Normality of $1N{H}_{3}P{O}_4$ is $1$N.

Therefore, the normality of $6g$ of $NaOH$ per $100mL$ is highest.

Hence, option C is correct.