JEE Mains · Chemistry · STD 11 - 7. Redox reactions
\(20\,mL\) of \(0.02\,M\) hypo solution is used for the titration of \(10\,mL\) of copper sulphate solution, in the presence of excess of \(KI\) using starch as an indicator. The molarity of \(Cu ^{2+}\) is found to be \(\times 10^{-2}\,M\)[nearest integer]Given : \(2 Cu ^{2+}+4 I ^{-} \rightarrow Cu _{2} I _{2}+ I _{2}\) \(I _{2}+2 S _{2} O _{3}{ }^{2-} \rightarrow 2 I ^{-}+ S _{4} O _{6}{ }^{2-}\)
- A \(3\)
- B \(4\)
- C \(5\)
- D \(6\)
Answer & Solution
Correct Answer
(B) \(4\)
Step-by-step Solution
Detailed explanation
\(n _{\text {eq. }}\) of \(I _{2}= n _{ eq }\) of \(Na _{2} S _{2} O _{3}=20 \times 0.002 \times 1\) \(2 \times n _{ mol }\) of \(I _{2}=0.4\) \(n _{ mol }\) of \(I _{2}=0.2 m mol\) \(n _{ mol }\) of \(Cu ^{+2}=0.2 \times 2 \times 10^{-3}\)…
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