Question

Which of the following solutions will have the lowest normality?

• A. $30\text{g}$ of $HCl$ in $500\text{mL}$ of solution
• B. $4.9\text{g}$ of $H_{2}S{O}_4$ in $500\text{mL}$ of solution
• C. $6.3\text{g}$ of $HN{O}_3$ in $250\text{mL}$ of solution
• D. $9.8\text{g}$ of $H_{3}P{O}_4$ in $500\text{mL}$ of solution

Answer 1

The correct option is B $4.9\text{g}$ of $H_{2}S{O}_4$ in $500\text{mL}$ of solution

(a) Gram equivalent of $HCl=\frac{\text{given mass}}{\text{equivalent mass}}=\frac{30}{36.5}=0.822$

$\text{Normality}=\frac{\text{gram equivalents of solute}}{\text{total volume of solution in mL}}\times 1000$

$\text{Normality}=\frac{0.822}{500}\times 1000=1.64N$

(b) Gram equivalent of $H_{2}S{O}_4=\frac{\text{given mass}}{\text{equivalent mass}}=\frac{4.9}{49}=0.1$

$\text{Normality}=\frac{0.1}{500}\times 1000=0.2N$

(c) Gram equivalent of $HN{O}_3=\frac{\text{given mass}}{\text{equivalent mass}}=\frac{6.3}{63}=0.1$

$\text{Normality}=\frac{0.1}{250}\times 1000=0.4N$

(d) 9.8 g of $H_{3}P{O}_4$ is present in 500 mL of solution
Gram equivalent of $H_{3}P{O}_4=\frac{\text{given mass}}{\text{equivalent mass}}=\frac{9.8}{\frac{98}{3}}=0.3$
Since, n – factor is $3$ for $H_{3}P{O}_4$
$\text{Normality}=\frac{0.3}{500}\times 1000=0.60N$