d
Given percentage of chlorine in an hydrocarbon \(= 3.55\%\) i,e.,

\(100\,g\) of chlorohydrocarbon has \(3.55\,g\) of chlorine.

\(1\,g\) of chlorohydrocarbon will have \(\frac{{3.55}}{{100}} = 0.0355\,g\) of chlorine.

Atomic wt. of \(Cl = 35.5\,g/mol\)

Number of moles of \(Cl\) \( = \frac{{0.0355\,g}}{{35.5\,g/mol}} = 0.001\) \(mole\)

Number of atoms of \(Cl = 0.001\,mole\) \(\times 6.023 \times 10^{23}\,mol^{-1}\)

\(= 6.023 \times 10^{20}\)