$320$ mg of a sample of magnesium having a coating of its oxide required $20$ ml of $0.1$ m hydrochloric acid for the complete neutralisation of the letter. The composition of the sample is:

87.5 %

Mg and

12.5 %

MgO

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12.5 %

Mg and

87.5 %

MgO

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80 %

Mg and

20 %

MgO

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20 %

Mg and

80 %

MgO

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Solution

The correct option is A $87.5 %$ Mg and $12.5 %$ MgO
Total mass of $M g$ + $M g O$ =320mg
As per the reaction,
$M g O$ + $2 H C l$ $\to$ $M g C l_{2}$ + $H_{2} O$
Here, Atomic weight of $M g O$ is 40gm and Atomic weight of $H C l$ is 73gm.
20ml of 0.1M $H C l$ contains 0.073gm of $H C l$
Therefore, 73gm of $H C l$ combines with 40gm of $M g O$
0.073gm of $H C l$ will combine with = $\frac{40 \times 0.073}{73} = 0.04 g = 40 m g$
Amount of $M g$ = 320-40=280mg
Amount of $M g O$ = 40mg
Percentage of $M g$ = $\frac{280 \times 100}{320} =$ 87.5%
Percentage of $M g O$ = $\frac{40 \times 100}{320} =$ 12.5%

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$M g + H_{2} O \to A + H_{2}$

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C O_{2}

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