Q. Volume of 0.1 M $K_{2}Cr_{2}O_{7}$ required to oxidize 35 ml of 0.5 M $FeSO_{4}$ solution is

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A

29.2 ml

B

145 ml

C

175 ml

D

58.9 ml

Solution:

$K_{2}Cr_{2}O_{7}+6FeSO_{4}+7H_{2}SO_{4} \rightarrow K_{2}SO_{4}+Cr_{2}\left(\right.SO_{4}\left(\left.\right)_{3}+3Fe_{2}\left(\right.SO_{4}\left(\left.\right)_{3}+7H_{2}O$
For $K_{2}Cr_{2}O_{7},M=0.1M,n_{1}=1,V_{1}=?$
For $FeSO_{4},M=0.5M,n_{2}=6,V_{2}=35$ ml
$\frac{M_{1} V_{1}}{n_{1}}=\frac{M_{2} V_{2}}{n_{2}}$
$V_{1}=29.2$ ml

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