Question

What is the equivalent mass of $H_{2}S{O}_4$ for the given reaction?
(Molar mass is given as ‘M’)
$H_{2}S{O}_4+NaOH\to NaHS{O}_4+H_{2}O$

• A. $\frac{M}{2}$
• B. $\frac{M}{1}$
• C. $\frac{M}{3}$
• D. $\frac{M}{4}$

Answer 1

The correct option is B $\frac{M}{1}$

n-factor for $H_{2}S{O}_4$ in the reaction is 1, as the number of replaceable hydrogen ions of the acid for this reaction is only one.
Equivalent mass = $\frac{\text{Molar mass}}{\text{n-factor}}=\frac{M}{1}$