Question

Volume of 0.1 M $K_2Cr_2O_7$ required to oxidise 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

$K_{2}Cr_{2}O_{7}+4H_{2}SO_{4} \rightarrow K_{2}SO_{4}+Cr_{2}\left(SO_{4}\right)_{3}+4H_{2}O+3\left[O\right]$ $2FeSO_{4}+H_{2}SO_{4}+\left[O\right] \rightarrow Fe_{2}\left(SO_{4}\right)_{3}+H_{2}O \times3$ $\frac{K_{2}Cr_{2}O_{7}+7 H_{2}SO_{4}+6FeSO_{4} \rightarrow K_{2}SO_{4}+Cr_{2} \left(SO_{4}\right)_{3} +7H_{2}O+3Fe_{2}\left(SO_{4}\right)_{3}}{\frac{M_{1} V_{1}}{n_{1}}=\frac{M_{2}V_{2}}{n_{2}}}$ $(K_{2}Cr_{2}O_{7}) (FeSO_{4})$ $\frac{0.1\times V_{1}}{1}=\frac{0.5\times35}{6\times0.1}$ $V_{1}=\frac{0.5\times35}{6\times0.1}=29.2\,mL$