Question

Volume of 0.1M $K_{2}C{r}_2{O}_7$ required to oxidize 35 mL of 0.5 M FeSO4 solution is:

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

Answer 1

The correct option is A

29.2 mL

$n_f$ of $K_{2}C{r}_2{O}_7=6$

$n_f$ of $FeS{O}_4=1$

Equivalents of $K_{2}C{r}_2{O}_7$ = Equivalents of $FeS{O}_4$

$(M\times V\times n_f{)}_{K_{2}C{r}_2{O}_7}=(M\times V\times n_f{)}_{FeS{O}_4}$

$0.1\times V\times 6=0.5\times 35\times 1$

$V=\frac{35}{0.1\times 6}=29.2mL$