In electrolytic reduction of a nitroarene with 50 - current efficiency- 20-50 g of the compound is reduced by 2 - 96500 C of electric charge- The molar mass of the compound is-A- 123-00 gB- 61-50 gC- 10-25 gD- 20-50 g

The correct option is A 123.00 g Ar NO_2 + 6H^+ + 6e^ → Ar NH_2 + 2H_2O W = z Q η = molar mass × 2 × 96500/6 × 96500×50/100⇒ molar mass = 6W = 6 × 20.50 ...

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Standard XII

Chemistry

Faraday's Second Law

In electrolyt...

Question

In electrolytic reduction of a nitroarene with 50% current efficiency, 20.50 g of the compound is reduced by $2 \times 96500$ C of electric charge. The molar mass of the compound is :

20.50 g

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10.25 g

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123.00 g

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61.50 g

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Solution

The correct option is C
123.00 g

$A r - N O_{2} + 6 H^{+} + 6 e^{-} \to A r - N H_{2} + 2 H_{2} O W = z Q η = \frac{m o l a r m a s s \times 2 \times 96500}{6 \times 96500} \times \frac{50}{100} \Rightarrow m o l a r m a s s = 6 W = 6 \times 20.50 g = 123 g$

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In electrolytic reduction of a nitroarene with 50% current efficiency, 20.50 g of the compound is reduced by 2×96500 C of electric charge. The molar mass of the compound is :

The correct option is C 123.00 g Ar−NO2+6H++6e−→Ar−NH2+2H2OW=zQη=molar mass×2×965006×96500×50100⇒molarmass=6W=6×20.50g=123g

In electrolytic reduction of a nitroarene with 50% current efficiency, 20.50 g of the compound is reduced by $2 \times 96500$ C of electric charge. The molar mass of the compound is :

The correct option is C
123.00 g

$A r - N O_{2} + 6 H^{+} + 6 e^{-} \to A r - N H_{2} + 2 H_{2} O W = z Q η = \frac{m o l a r m a s s \times 2 \times 96500}{6 \times 96500} \times \frac{50}{100} \Rightarrow m o l a r m a s s = 6 W = 6 \times 20.50 g = 123 g$