Volume of $0.1 M K_{2} C r_{2} O_{7}$ required to oxidise $35 m L$ of $0.5 M F e S O_{4}$ solution is

29.2 m L

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17.5 m L

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175 m L

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145 m L

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Solution

The correct option is A $29.2 m L$

$K_{2} C r_{2} O_{7} + 6 F e S O_{4} + 7 H_{2} S O_{4} \to C r_{2} (S O_{4})_{3} + 3 F e_{2} (S O_{4})_{3} + K_{2} S O_{4} + 7 H_{2} O$

$M_{1}, V_{1}, n_{1}$ are molarity ,volume and number of moles of $K_{2} C r_{2} O_{7}$

$M_{2}, V_{2}, n_{2}$ are molarity ,volume and number of moles of $F e S O_{4}$

$\frac{M_{1} V_{1}}{n_{1}} = \frac{M_{2} V_{2}}{n_{2}}$

So $V_{1} = \frac{M_{2} V_{2}}{n_{2}} \times \frac{n_{1}}{M_{1}}$

$V_{1} = \frac{0.5 \times 35}{6} \times \frac{1}{0.1} = 29.2$

Hence, the correct option is $A$

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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:

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