Volume of 0.1M $K_{2} {C r}_{2} O_{7}$ required to oxidise 35ml of 0.5M ${F e S O}_{4}$ solution is:

35 m l

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29 m l

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17.5 m l

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175 m l

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Solution

The correct option is B $29 m l$
$K_{2} C r_{2} O_{7} + 6 F e S O_{4} + 7 H_{2} S O_{4} \to C r_{2} (S O_{4})_{3} + 3 F e_{2} (S O_{4})_{3} + K_{2} S O_{4} + 7 H_{2} O$

Valency factor of $K_{2} C r_{2} O_{7} = 6$

And the valency factor of $F e S O_{4} = 1$
Now, $E q_{K_{2} C r_{2} O_{7}} = E q_{F e S O_{4}}$
$6 \times 0.1 \times V = 35 \times 0.5$

$\Rightarrow V = 175 / 6 = 29.1 m l$

So, the correct option is $B$

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Similar questions

Volume of 0.1M $K_{2} C r_{2} O_{7}$ required to oxidize 35 mL of 0.5 M FeSO4 solution is:

0.1 M K_{2} C r_{2} O_{7}

35 m L

0.5 M F e S O_{4}

H_{2} O_{2}

K_{2} {C r}_{2} O_{7}

{S O}_{2}

K_{2} {C r}_{2} O_{7}

C r

K_{2} {C r}_{2} O_{7}

K_{2} {C r}_{2} O_{7} = 294.0 g

m o l

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