KCET · Chemistry · Electrochemistry
An aqueous solution containing \(6.5 \mathrm{~g}\) of \(\mathrm{NaCl}\) of \(90 \%\) purity was subjected to electrolysis. After the complete electrolysis, the solution was evaporated to get solid \(\mathrm{NaOH}\). The volume of \(1 \mathrm{M}\) acetic acid required to neutralise \(\mathrm{NaOH}\) obtained above is
- A \(1000 \mathrm{~cm}^{3}\)
- B \(2000 \mathrm{~cm}^{3}\)
- C \(100 \mathrm{~cm}^{3}\)
- D \(200 \mathrm{~cm}^{3}\)
Answer & Solution
Correct Answer
(C) \(100 \mathrm{~cm}^{3}\)
Step-by-step Solution
Detailed explanation
Weight of pure \(\mathrm{NaCl}=6.5 \times 0.9=5.85 \mathrm{~g}\)
No. of equivalence of \(\mathrm{NaCl}=\frac{5.85}{58.5}=0.1\)
No. of equivalence of \(\mathrm{NaOH}\) obtained \(=0.1\)
Volume of \(1 \mathrm{M}\) acetic acid required for the neutralisation of \(\mathrm{NaOH}=\frac{0.1 \times 1000}{1}\)
\[
=100 \mathrm{~cm}^{3}
\]
No. of equivalence of \(\mathrm{NaCl}=\frac{5.85}{58.5}=0.1\)
No. of equivalence of \(\mathrm{NaOH}\) obtained \(=0.1\)
Volume of \(1 \mathrm{M}\) acetic acid required for the neutralisation of \(\mathrm{NaOH}=\frac{0.1 \times 1000}{1}\)
\[
=100 \mathrm{~cm}^{3}
\]
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