JEE MAIN - Chemistry (2022 - 28th June Evening Shift - No. 21)
0.01 M KMnO4 solution was added to 20.0 mL of 0.05 M Mohr's salt solution through a burette. The initial reading of 50 mL burette is zero. The volume of KMnO4 solution left in the burette after the end point is _____________ mL. (nearest integer)
Answer
30
Explanation
Meq of oxidizing agent $=$ Meq of reducing agent
$\left(\mathrm{M} \times \mathrm{V} \times \mathrm{n}_{\text{F}}\right)_{\mathrm{KMnO}_{4}}=\left(\mathrm{M} \times \mathrm{V} \times \mathrm{n}_{\text{F}}\right)_{\text {Mohr's salt }}$
$0.01 \times 20 \times 5=0.05 \times V \times 1$
Volume required $=20 ~ \mathrm{ml}$
Since initial volume of $\mathrm{KMnO}_{4}$ in burette is $50 ~\mathrm{ml}$.
Hence volume of $\mathrm{KMnO}_{4}$ left in the burette after end point is $30 ~\mathrm{ml}$.
$\left(\mathrm{M} \times \mathrm{V} \times \mathrm{n}_{\text{F}}\right)_{\mathrm{KMnO}_{4}}=\left(\mathrm{M} \times \mathrm{V} \times \mathrm{n}_{\text{F}}\right)_{\text {Mohr's salt }}$
$0.01 \times 20 \times 5=0.05 \times V \times 1$
Volume required $=20 ~ \mathrm{ml}$
Since initial volume of $\mathrm{KMnO}_{4}$ in burette is $50 ~\mathrm{ml}$.
Hence volume of $\mathrm{KMnO}_{4}$ left in the burette after end point is $30 ~\mathrm{ml}$.
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