Question

$32\, g$ of a sample of $FeSO _{4} \cdot 7 H _{2} O$ were dissolved in dilute sulphuric acid and water and its volume was made up to 1litre. $25\, mL$ of this solution required $20 \, mL$ of $0.02\, M\, KMnO _{4}$ solution for complete oxidation. Calculate the mass $\%$ of $FeSO _{4} \cdot 7 H _{2} O$ in the sample.

• A. $34.75$
• B. $69.5$
• C. $89.5$
• D. None of these

Answer 1

Answer: (B)

$MnO _{4}+5 Fe ^{2+}+8 H ^{+} \rightarrow Mn ^{2+}+5 Fe ^{3+}+4 H _{2} O$
$n_{f}\left( KMnO _{4}\right)=7-2=5$
$n_{f}\left( FeSO _{4}\right)=3-2=1$
$m$-eq. of $FeSO _{4} \cdot 7 H _{2} O$ in 1 litre $= m$ - eq. of $KMnO _{4}$
$\Rightarrow 20 \times 0.02 \times 5 \times\left(\frac{1000}{25}\right)=\frac{W}{278} \times 1 \times 1000$
[Mass of $\left.FeSO _{4} \cdot 7 H _{2} O =278\, g / mol \right]$
$\Rightarrow W =22.24$
Mass $\%$ of $FeSO _{4} \cdot 7 H _{2} O$ in given sample
$=\frac{22.24}{32} \times 100=69.5$