$3.92 \mathrm{~g}$ of ferrous ammonium sulphate crystals are dissolved in $100 \mathrm{~mL}$ of water. $20…

$3.92 \mathrm{~g}$ of ferrous ammonium sulphate crystals are dissolved in $100 \mathrm{~mL}$ of water. $20 \mathrm{~mL}$ of this solution requires $18 \mathrm{~mL}$ of potassium permaganate during titration for complete oxidation. The weight of $\mathrm{KMnO}_{4}$ present in one litre of the solution of
  1. $3.47 \mathrm{~g}$
  2. $12.38 \mathrm{~g}$
  3. $1.23 \mathrm{~g}$
  4. $34.76 \mathrm{~g}$

Solution

Normality of ferrous amm. sulphate
$=\frac{3.92 \times 1000}{392 \times 100}=0.1($ Eq. wt of FAS is 392$)$
$\mathrm{N}_{1} \mathrm{~V}_{1}=\mathrm{N}_{2} \mathrm{~V}_{2}$
$20 \times 0.1=18 \times \mathrm{N}_{2} \quad \mathrm{~N}_{2}=0.111$
g ev. of $\mathrm{KMnO}_{4}=31.6 \mathrm{~g}$
$0.111 \mathrm{~g}$ ev. of $\mathrm{KMnO}_{4}=31.6 \times 0.111=3.5 \mathrm{~g}$.

Asked in: JEE-TOPICTESTS-CHEMISTRY

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