TS EAMCET · Chemistry · Redox Reactions
\(25 \mathrm{~mL}\) of \(0.1 \mathrm{~N} \mathrm{NaOH}\) solution neutralises \(12.5 \mathrm{~mL}\) of \(\mathrm{HCl}\) solution. The amount of water needed to convert \(500 \mathrm{~mL}\) of such \(\mathrm{HCl}\) solution to \(0.1 \mathrm{~N}\) is
- A \(555 \mathrm{~mL}\)
- B \(500 \mathrm{~mL}\)e
- C \(50 \mathrm{~mL}\)
- D \(55.5 \mathrm{~mL}\)
Answer & Solution
Correct Answer
(B) \(500 \mathrm{~mL}\)e
Step-by-step Solution
Detailed explanation
\(\because\) Concentration of \(\mathrm{NaOH}\left(N_1\right)=0.1 \mathrm{~N}\) Volume of \(\mathrm{NaOH}\) used \(\left(V_1\right)=25 \mathrm{~mL}\) Concentration of \(\mathrm{HCl}\left(N_2\right)=0.1 \mathrm{~N}\) Volume of \(\mathrm{HCl}\left(V_2\right)\) used \(=V_2\) and…
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