Question

$200mL$ of $3NHCl$ were mixed with $200mL$ of $6N{H}_{2}S{O}_4$ solution. The final normality of $H_{2}S{O}_4$ in the resultant solution will be

• A. $9N$
• B. $3N$
• C. $6N$
• D. $2N$

Answer 1

The correct option is B $3N$

$N_1,V_1$ are normality and volume of $HCl$
$N_2,V_2$ are normality and volume of $H_{2}S{O}_4$ respectively
$N=\frac{N_1{V}_1+N_2{V}_2}{V_1+V_2}$
$N=\frac{200\times 3+200\times 6}{200+300}=3.6$N
Hence option B is correct