Question

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Answer 1

Answer:

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An unknown chlorohydrocarbon has 3.55% of chlorine.

100 g of chlorohydrocarbon has 3.55 g of chlorine.

1 g of chlorohydrocarbon will have $3.55 \times \frac{1}{100}$

$= 0.0355g \: of \: chlorine$

Atomic wt. of Cl=35.5 g/mol

Atomic wt. of Cl=35.5 g/mol

Number of moles of Cl= $\frac{0.0355g}{35.5g</em><em>/</em><em>mol</em><em>}$

=0.001mol

=0.001molNumber of atoms of Cl=0.001mol$\times 6.023 \times {10}^{23} {mol}^{ - 1}$

$= 6.023 \times {10}^{20}$

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