10 mL of a \(𝐻_{2}𝑆𝑂_{4}\) solution is diluted to 100 mL. 25 mL of this diluted solution is mixed with 50 mL of a 0.5 N NaOH solution. The resulting solution requires 0.265 g of \(𝑁𝑎_{2}𝐶𝑂_{3}\) for complete neutralisation. The normality of the original \(𝐻_{2}𝑆𝑂_{4}\) solution is

Calculating the normality of \(𝑯_{𝟐}𝑺𝑶_{𝟒}\) solution used to react with \(𝑵𝒂𝑶𝑯\), and neutralise \(𝑵𝒂_{𝟐}𝑪𝑶_{𝟑}\)

Assuming, Normality of \(𝐻_{2}𝑆𝑂_{4} = 𝑁_2\)
\(𝑁𝑢𝑚𝑏𝑒𝑟 ~𝑜𝑓 ~𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠 = 𝑀𝑜𝑙𝑒𝑠 × 𝑛 − 𝑓𝑎𝑐𝑡𝑜r\)
\(\dfrac{Mass}{Molar~ mass}\times~n-factor\)

Milliequivalents of \(𝐻_{2}𝑆𝑂_{4}\) = Milliequivalents of \(𝑁𝑎𝑂𝐻 + Milliequivalents ~of~ 𝑁𝑎_{2}𝐶𝑂_{3}\)

\(𝑁_{2}\times 𝑉_{𝐻_2𝑆𝑂_4}\)

= \(N_{N_aOH}\times_{V_{NaOH}}+\dfrac{m_{Na_2CO_3}}{M_{Na_2CO_3}}\times~n-factor\times1000\)

\(N_{2}\times25=0.5\times50+\frac{0.265}{106}\times2\times1000\)

\(N_{2}=\dfrac{25+5}{25}=\dfrac{30}{25}=1.2~N\)

Calculating the normality of the original \(𝑯_{𝟐}𝑺𝑶_{𝟒}\) solution

For dilution process:

\(𝑁_{1}𝑉_{1} = 𝑁_{2}𝑉_{2}\)

\(\Rightarrow 𝑁_{1}\times 10 = 𝑁_{2}\times 100\)

\(N_{2}=\dfrac{1.2\times100}{10}=12~N\)

Hence, the normality of the original \(𝐻_{2}𝑆𝑂_{4}\) solution is \(12 ~𝑁\).

So, option (A) is the correct answer.