$3.92 \mathrm{~g}$ of ferrous ammonium sulphate crystals are dissolved in $100 \mathrm{~mL}$ of water. $20…
- $3.47 \mathrm{~g}$
- $12.38 \mathrm{~g}$
- $1.23 \mathrm{~g}$
- $34.76 \mathrm{~g}$
Solution
$=\frac{3.92 \times 1000}{392 \times 100}=0.1($ 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
Practice more SOME BASIC CONCEPTS OF CHEMISTRY questions on Aicharya