Question

5.5 gm of a mixture of $FeS{O}_4.7H_{2}O$ and $F{e}_2(S{O}_4{)}_3.$ $9H_{2}O$ requires 5.4 ml of 0.1 N $KMn{O}_4$ solution for complete oxidation. Calculate the number of gram moles of hydrated ferric sulphate in the mixture?

• A. $9.5\times {10}^{-2}$ mol
• B. $9.5\times {10}^{-3}$ mol
• C. $9.5\times {10}^3$ mol
• D. $9.5\times {10}^2$ mol

Answer 1

The correct option is B $9.5\times {10}^{-3}$ mol

$F{e}^{+2}\to F{e}^{+3}+e^{-}$
Molar mass of $FeS{O}_4.7H_{2}O=278g/mol$
$\therefore$ Equivalent mass of $FeS{O}_4.7H_{2}O=\frac{278}{1}=278$
$\because 5.4mLof0.1NKMn{O}_4\equiv 5.4mLof0.1NFeS{O}_4.7H_{2}O$
$\Rightarrow \text{Amount of}FeS{O}_4.7H_{2}O=\frac{0.1\times 278}{1000}\times 5.4=0.15g$
$\Rightarrow \text{Amount of}F{e}_2(S{O}_4{)}_3.9H_{2}O=(5.5-0.15)g$
$\therefore \text{number of gram moles of}F{e}_2(S{O}_4{)}_3.9H_{2}O=\frac{\text{mass}}{\text{molar mass}}=\frac{5.35}{562}=9.52\times {10}^{-3}mol$