Question

Volume of $3\%$ solution of sodium carbonate necessary to neutralise a litre of $0.1N$ sulphuric acid

• A. 176.66 ml
• B. 156.6 ml
• C. 116.0 ml
• D. 196.1 ml

Answer 1

Answer: (A)

[a] Normality of $3\%N{a}_{2}C{O}_3$
$N=\frac{3\times 100}{53\times 100}=0.566N$
For $H_{2}S{O}_4$ sol. $N_1=0.1,V_1=100mL$
For $N{a}_{2}C{O}_3$ sol. $N_2=\mathrm{0.566.}{N}_2$ ?
Now apply $N_1{V}_1=N_2{V}_2$
$V_2=\frac{N_1{V}_1}{N_2}=\frac{0.1\times 100mL}{0.566}$
$=176.66mL$