Question

Volume of $0.1\, M\, K _{2} Cr _{2} O _{7}$ required to oxidize $35\, mL$ of $0.5\, M\, FeSO _{4}$ solution is

• A. 29.2 mL
• B. 17.5 mL
• C. 175 mL
• D. 145 mL

Answer 1

Answer: (A)

29.2 mL

Answer 2

$Cr _{2} O _{7}^{2-}+6 Fe ^{2+}+14 H ^{+} \rightarrow 6 Fe ^{3+}$
$+2 Cr ^{3+}+7 H _{2} O$ Hence, $1$ mol of $Cr _{2} O _{7}^{2-}=6$
mole of $Fe ^{2+}$
$\frac{M_{1} V_{1}}{1}=\frac{M_{2} V_{2}}{6}$
$V_{1}=\frac{0.1 \times V_{1}}{6 \times 0.1}$
$\frac{0.1 \times V_{1}}{1}=\frac{0.5 \times 35}{6}$
$V_{1}=29.2\, mL$