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Volume of $3 \%$ solution of sodium carbonate necessary to neutralise a litre of $0.1 \mathrm{~N}$ sulphuric acid
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The correct answer is:
$176.66 \mathrm{ml}$
Normality of $3 \% \mathrm{Na}_{2} \mathrm{CO}_{3}$
$\mathrm{N}=\frac{3 \times 1000}{53 \times 100}=0.566 \mathrm{~N}$
For $\mathrm{H}_{2} \mathrm{SO}_{4}$ sol. $\mathrm{N}_{1}=0.1, \mathrm{~V}_{1}=100 \mathrm{~mL}$
For $\mathrm{Na}_{2} \mathrm{CO}_{3}$ sol. $\mathrm{N}_{2}=0.566$
Now apply $\mathrm{N}_{1} \mathrm{~V}_{1}=\mathrm{N}_{2} \mathrm{~V}_{2}$
$\mathrm{V}_{2}=\frac{\mathrm{N}_{1} \mathrm{~V}_{1}}{\mathrm{~N}_{2}}=\frac{0.1 \times 1000 \mathrm{~mL}}{0.566}=176.66 \mathrm{~mL}$
$\mathrm{N}=\frac{3 \times 1000}{53 \times 100}=0.566 \mathrm{~N}$
For $\mathrm{H}_{2} \mathrm{SO}_{4}$ sol. $\mathrm{N}_{1}=0.1, \mathrm{~V}_{1}=100 \mathrm{~mL}$
For $\mathrm{Na}_{2} \mathrm{CO}_{3}$ sol. $\mathrm{N}_{2}=0.566$
Now apply $\mathrm{N}_{1} \mathrm{~V}_{1}=\mathrm{N}_{2} \mathrm{~V}_{2}$
$\mathrm{V}_{2}=\frac{\mathrm{N}_{1} \mathrm{~V}_{1}}{\mathrm{~N}_{2}}=\frac{0.1 \times 1000 \mathrm{~mL}}{0.566}=176.66 \mathrm{~mL}$
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