If $2.77 \times 10^{25}$ electrons are there in a $M g C l_{2}$ sample, then what is the mass of $M g C l_{2}$ present?

95 g

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190 g

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45 g

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110 g

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Solution

The correct option is A $95 g$
The number of electrons in one molecule of $M g C l_{2} = (12 + (17 \times 2)) = 46$
So, the number of electrons in one mol of $M g C l_{2} = 46 \times 6.022 \times 10^{23}$
Therefore,
$46 \times 6.022 \times 10^{23}$ electrons are contained in 1 mol of $M g C l_{2}$
Hence $2.77 \times 10^{25}$ electrons will be contained in = $\frac{2.77 \times 10^{25}}{46 \times 6.022 \times 10^{23}}$ mol of $M g C l_{2}$
= 0.999 mol of $M g C l_{2}$
Thus, the required mass of $M g C l_{2} = (24 + 2 \times 35.5) \times 0.999 g$
$= 94.905 g ≅ 95 g$

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If $2.77 \times 10^{25}$ electrons are there in a $M g C l_{2}$ sample, then what is the mass of $M g C l_{2}$ present?