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  4. An unknown chlorohydrocarbon has 3.55 % of chlorine. If each molecule of the hydrocarbon has one chlorine atom only; chlorine atoms present in 1 g of chlorohydrocarbon are: (Atomic wt. of Cl=35.5 u; Avogadro constant = 6.023 × 1023 mol-1)

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Q. An unknown chlorohydrocarbon has $3.55\%$ of chlorine. If each molecule of the hydrocarbon has one chlorine atom only; chlorine atoms present in $1\, g$ of chlorohydrocarbon are :
(Atomic wt. of $Cl=35.5\, u;$ Avogadro constant = $6.023 \times 10^{23} \, mol^{-1}$)

JEE MainJEE Main 2018Some Basic Concepts of Chemistry

Solution:

$3.55 \%$ chlorine indicates that $3.55\, g$ of chlorine is present in $100\, g$ of chlorohydrocarbon.

Therefore, $1\, g$ of chlorohydrocarbon contains $3.55 \times \frac{1}{100}=0.0355\, g$ of chlorine

Number of moles of chlorine can be calculated as

$\frac{0.0355\, g }{35.5\, g\, mol ^{-1}}=0.001 mol$

Number of moles of $Cl$ atoms $=0.001 mol \times 6.023 \times 10^{23} mol ^{-1}=6.023 \times 10^{20}$