The normality of $\mathrm{H}_{2} \mathrm{SO}_{4}$ in the solution obtained on mixing $100 \mathrm{~mL}$ of $0.1 \,\mathrm{M} \,\mathrm{H}_{2} \mathrm{SO}_{4}$ with $50 \mathrm{~mL}$ of $0.1 \,\mathrm{M}\, \mathrm{NaOH}$ is _______________ $\times 10^{-1} \mathrm{~N}$. (Nearest Integer)
Answer (integer)
1
Solution
No. of equivalents of $\mathrm{H}_2 \mathrm{SO}_4=100 \times 0.1 \times 2=20$<br/><br/>
No. of equivalents of $\mathrm{NaOH}=50 \times 0.1=5$<br/><br/>
No. of equivalents of $\mathrm{H}_2 \mathrm{SO}_4$ left $=20-5=15$<br/><br/>
$$
\begin{aligned}
&\Rightarrow 150 \times \mathrm{x}=15 \\\\
&\mathrm{x}=\frac{1}{10}=0.1 \mathrm{~N}=1 \times 10^{-1} \mathrm{~N}
\end{aligned}
$$
About this question
Subject: Chemistry · Chapter: Redox Reactions · Topic: Oxidation States
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