JEE MAIN - Chemistry (2022 - 27th July Evening Shift - No. 15)
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
1
Explanation
No. of equivalents of $\mathrm{H}_2 \mathrm{SO}_4=100 \times 0.1 \times 2=20$
No. of equivalents of $\mathrm{NaOH}=50 \times 0.1=5$
No. of equivalents of $\mathrm{H}_2 \mathrm{SO}_4$ left $=20-5=15$
$$ \begin{aligned} &\Rightarrow 150 \times \mathrm{x}=15 \\\\ &\mathrm{x}=\frac{1}{10}=0.1 \mathrm{~N}=1 \times 10^{-1} \mathrm{~N} \end{aligned} $$
No. of equivalents of $\mathrm{NaOH}=50 \times 0.1=5$
No. of equivalents of $\mathrm{H}_2 \mathrm{SO}_4$ left $=20-5=15$
$$ \begin{aligned} &\Rightarrow 150 \times \mathrm{x}=15 \\\\ &\mathrm{x}=\frac{1}{10}=0.1 \mathrm{~N}=1 \times 10^{-1} \mathrm{~N} \end{aligned} $$
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