When a quantity of electricity is passed through $C u S O_{4}$ solution, $0.16 g$ of copper gets deposited. If the same quantity of electricity is passed through acidulated water, then the volume of $H_{2}$ liberated at STP will be: (given atomic weight of $C u = 64$ ).

4 c m^{3}

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56 c m^{3}

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604 c m^{3}

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8 c m^{3}

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Solution

The correct option is B $56 c m^{3}$
Number of equivalents of copper deposited $= \frac{0.16}{32} = 0.005$
Volume of $H_{2}$ gas at $S T P = 11.2 \times 0.005$
$= 0.056 l i t r e = 56 c m^{3}$ .

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When a quantity of electricity is passed through CuSO4 solution, 0.16 g of copper gets deposited. If the same quantity of electricity is passed through acidulated water, then the volume of H2 liberated at STP will be: (given atomic weight of Cu=64).

The correct option is B 56 cm3Number of equivalents of copper deposited =0.1632=0.005Volume of H2 gas at STP=11.2×0.005=0.056 litre=56 cm3.

When a quantity of electricity is passed through $C u S O_{4}$ solution, $0.16 g$ of copper gets deposited. If the same quantity of electricity is passed through acidulated water, then the volume of $H_{2}$ liberated at STP will be: (given atomic weight of $C u = 64$ ).

The correct option is B $56 c m^{3}$
Number of equivalents of copper deposited $= \frac{0.16}{32} = 0.005$
Volume of $H_{2}$ gas at $S T P = 11.2 \times 0.005$
$= 0.056 l i t r e = 56 c m^{3}$ .