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Standard XII

Chemistry

Electrolysis and Electrolytes

6.5 g Mg de...

Question

6.5 g $M g$ deposits on electrolysis of $M g C l_{2}$ (fused) then what will be the volume of chlorine gas at STP discharged at electrode?

6.067 L

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3.033 L

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9.1 L

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12.133 L

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Solution

The correct option is A 6.067 L
$1$ mole of $M g C l_{2}$ gives $1 m o l$ of $M g (24 g)$ & one mole of $C l_{2}$
$24 g$ of $M g = 22.4 d m^{3}$ of $C l_{2}$ gas at STP
$∴ 6.5 g$ of $M g = \frac{22.4 \times 6.5}{24}$
$= 6.067 L$ of $C l_{2}$ gas

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Similar questions

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6.5 g Mg deposits on electrolysis of MgCl2 (fused) then what will be the volume of chlorine gas at STP discharged at electrode?

The correct option is A 6.067 L1 mole of MgCl2 gives 1mol of Mg(24g) & one mole of Cl224g of Mg=22.4dm3 of Cl2 gas at STP∴6.5g of Mg=22.4×6.524=6.067L of Cl2 gas

6.5 g $M g$ deposits on electrolysis of $M g C l_{2}$ (fused) then what will be the volume of chlorine gas at STP discharged at electrode?

The correct option is A 6.067 L
$1$ mole of $M g C l_{2}$ gives $1 m o l$ of $M g (24 g)$ & one mole of $C l_{2}$
$24 g$ of $M g = 22.4 d m^{3}$ of $C l_{2}$ gas at STP
$∴ 6.5 g$ of $M g = \frac{22.4 \times 6.5}{24}$
$= 6.067 L$ of $C l_{2}$ gas