Question

Volume of 0.1 M $K_{2}C{r}_2{O}_7$ required to oxidize 35 ml of 0.5 M $FeS{O}_4$ solution is

• A. 29.2 ml
• B. 145 ml
• C. 175 ml
• D. 58.9 ml

Answer 1

Answer: (A)

$K_{2}C{r}_2{O}_7+6FeS{O}_4+7H_{2}S{O}_4\to K_{2}S{O}_4+C{r}_2{\left(S{O}_4\right)}_3$ $+3F{e}_2{\left(S{O}_4\right)}_3+7H_{2}O$

For $K_{2}C{r}_2{O}_7,M=0.1M,n_1=1,V_1=?$

For $FeS{O}_4,M=0.5M,n_2=6,V_2=35$ ml

$\frac{M_1{V}_1}{n_1}=\frac{M_2{n}_2}{n_2}$

$V_1=29.2$ ml