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$100 \mathrm{~g}$ of a mixture of $\mathrm{NaOH}$ and $\mathrm{Na}_2 \mathrm{SO}_4$ is neutralised by $100 \mathrm{~mL}$ of $0.5 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$. What is the amount of $\mathrm{Na}_2 \mathrm{SO}_4$ present in the mixture?
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Verified Answer
The correct answer is:
96 g
Reaction of neutralisation
$$
\mathrm{NaOH}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{Na}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}
$$
Number of equivalent of $\mathrm{NaOH}=$ Number of equivalent of $\mathrm{H}_2 \mathrm{SO}_4$
Number of equivalent mol $\mathrm{H}_2 \mathrm{SO}_4=100 \times 0.5$
$$
=50 \mathrm{~m} \mathrm{~mol}
$$
$50 \times n$-factor $\Rightarrow 50 \times 2=100$ milli equivalent.
Number of Eq. of $\mathrm{NaOH}$
$$
\begin{aligned}
& =N \times 100 \mathrm{~mL}(N=\text { Normality }) \\
& =100 \mathrm{~N}
\end{aligned}
$$
Number of Eq. of $\mathrm{NaOH}=$ Number of Eq. of $\mathrm{H}_2 \mathrm{SO}_4$
$$
\begin{aligned}
100 N & =100 \\
N & =1
\end{aligned}
$$
Number of mole of $\mathrm{NaOH}$ in $1000 \mathrm{~mL}=1$
Number of mole of $\mathrm{NaOH}$ in $100 \mathrm{~mL}=0.1$
Mass of $\mathrm{NaOH}$ in $100 \mathrm{~mL}=0.1 \times 40=4 \mathrm{~g}$
Mass of $\mathrm{Na}_2 \mathrm{SO}_4=100-4=96 \mathrm{~g}$
$$
\mathrm{NaOH}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{Na}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}
$$
Number of equivalent of $\mathrm{NaOH}=$ Number of equivalent of $\mathrm{H}_2 \mathrm{SO}_4$
Number of equivalent mol $\mathrm{H}_2 \mathrm{SO}_4=100 \times 0.5$
$$
=50 \mathrm{~m} \mathrm{~mol}
$$
$50 \times n$-factor $\Rightarrow 50 \times 2=100$ milli equivalent.
Number of Eq. of $\mathrm{NaOH}$
$$
\begin{aligned}
& =N \times 100 \mathrm{~mL}(N=\text { Normality }) \\
& =100 \mathrm{~N}
\end{aligned}
$$
Number of Eq. of $\mathrm{NaOH}=$ Number of Eq. of $\mathrm{H}_2 \mathrm{SO}_4$
$$
\begin{aligned}
100 N & =100 \\
N & =1
\end{aligned}
$$
Number of mole of $\mathrm{NaOH}$ in $1000 \mathrm{~mL}=1$
Number of mole of $\mathrm{NaOH}$ in $100 \mathrm{~mL}=0.1$
Mass of $\mathrm{NaOH}$ in $100 \mathrm{~mL}=0.1 \times 40=4 \mathrm{~g}$
Mass of $\mathrm{Na}_2 \mathrm{SO}_4=100-4=96 \mathrm{~g}$
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