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Conductance and Conductivity

50 cm 3 of ...

Question

$50 {c m}^{3}$ of $0.04 M K_{2} {C r}_{2} O_{7}$ in acidic medium oxidizes a sample of $H_{2} S$ gas to sulphur. The volume of $0.03 M K M n O_{4}$ required to oxidize the same amount of $H_{2} S$ gas to sulphur, in acidic medium is:

80 {c m}^{3}

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120 {c m}^{3}

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60 {c m}^{3}

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90 {c m}^{3}

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Solution

The correct option is A $80 {c m}^{3}$
Equivalents of $K_{2} C r_{2} O_{7} =$ Equivalents of $H_{2} S =$ Equivalents of $K M n O_{4}$

$n_{f}$ of $K_{2} C r_{2} O_{7} = 6$
$n_{f}$ of $K M n O_{4} = 5$

But, we know that,

Equivalents $= n_{f} \times M o l a r i t y \times V o l u m e$

$\Rightarrow 6 \times 0.04 \times 50 = 5 \times 0.03 \times V_{K M n O_{4}}$

$V_{K M n O_{4}} = 80 c m^{3}$

Hence, the correct option is $A$

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Similar questions

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Q. In acidic medium, 100 ml of 1 M

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50cm3 of 0.04 M K2Cr2O7 in acidic medium oxidizes a sample of H2S gas to sulphur. The volume of 0.03 M KMnO4 required to oxidize the same amount of H2S gas to sulphur, in acidic medium is:

The correct option is A 80 cm3Equivalents of K2Cr2O7= Equivalents of H2S= Equivalents of KMnO4nf of K2Cr2O7=6nf of KMnO4=5But, we know that,Equivalents =nf×Molarity×Volume⇒6×0.04×50=5×0.03×VKMnO4VKMnO4=80 cm3Hence, the correct option is A

$50 {c m}^{3}$ of $0.04 M K_{2} {C r}_{2} O_{7}$ in acidic medium oxidizes a sample of $H_{2} S$ gas to sulphur. The volume of $0.03 M K M n O_{4}$ required to oxidize the same amount of $H_{2} S$ gas to sulphur, in acidic medium is: