Calculate the total number of electrons present in 1.6 g of methane.

6.02 \times 10^{22}

electrons

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6.02 \times 10^{32}

electrons

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6.02 \times 10^{23}

electrons

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6.02 \times 10^{20}

electrons

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Solution

The correct option is C $6.02 \times 10^{23}$ electrons
$16 g$ of Methane has $1 m o l$ of $C H_{4}$ .

$1 C H_{4}$ molecule has $10$ electrons.

Thus, $1.6 g$ has $0.1 m o l$ of $C H_{4}$ i.e., $(10 \times 0.1) m o l$ electrons $= 1 m o l$ electrons $= 6.02 \times 10^{23}$ electrons.

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K_{2} S O_{4} .

6.02 \times 10^{23}

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a \times 10^{23}

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N_{A} = 6.02 \times 10^{23}

C l = 35.5 g

1.6

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