22.7 𝑚𝐿 of (𝑁/10) \(𝑁𝑎_{2}𝐶𝑂_{3}\) solution neutralises 10.2 𝑚𝐿 of a dilute \(𝐻_{2}𝑆𝑂_{4}\) solution. The volume of water that must be added to 400 𝑚𝐿 of this \(𝐻_{2}𝑆𝑂_{4}\) solution in order to make it exactly 𝑁/10 is:

Calculating the normality of \(𝑯_{𝟐}𝑺𝑶_{𝟒}\) to neutralise \(𝑵𝒂_{𝟐}𝑪𝑶_{𝟑}\)
For the neutralisation reaction

\(𝐻_{2}𝑆𝑂_{4} + 𝑁𝑎_{2}𝐶𝑂_{3} → 𝑁𝑎_{2}𝑆𝑂_{4} + 𝐻_{2}
𝑂 + 𝐶𝑂_{2}\)
Equivalents of \(𝐻_{2}𝑆𝑂_{4} = Equivalents ~of~ 𝑁𝑎_{2}𝐶𝑂_{3}\)

\(N_{1}\times\dfrac{10.2}{1000}=\dfrac{1}{10}\times\dfrac{22.7}{1000}\)

\(N_{1}\times\dfrac{1\times22\times1000}{10.2\times10\times1000}=0.222549~N\)

Calculating the volume of water required for dilution

For dilution, \(𝑁_{1}𝑉_{1} = 𝑁_{2}𝑉_{2}\)

Where, \(𝑁_{1} = 0.222549 ~𝑁\)

\(N_{2}=\dfrac{N}{10}\)

\(𝑉_{1} = 400~ 𝑚𝐿\)

\(𝑉_{2} = (400 + 𝑥) 𝑚L\)

\(0.222549 ~𝑁\times400=\dfrac{N}{10}\times(400+x)\)

\(890.2 = (400 + 𝑥)\)

\(𝑥 = 890.2 – 400 = 490.2 𝑚L\)

Hence, 490.2 𝑚𝐿 of water should be added to 400 𝑚𝐿 of this \(𝐻_{2} 𝑆𝑂_{4}\) solution in order to make it exactly 𝑁/10.

So, option (A) is the correct answer