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$$^{-1}$$)

We are told that chlorine constitutes 3.55 % of the mass of the unknown chlorohydrocarbon. This means that, out of every 100 g of the substance, 3.55 g is chlorine. We wish to deal with just 1 g of the substance, so we scale the mass of chlorine proportionally.

For 1 g of the chlorohydrocarbon the mass of chlorine present is

$$\text{mass of Cl in 1 g} \;=\;\frac{3.55}{100}\times 1\ \text{g}\;=\;0.0355\ \text{g}.$$

Now we convert this mass of chlorine to moles. First, we state the molar-mass relation:

Formula: $$n = \frac{m}{M},$$ where $$n$$ is the number of moles, $$m$$ is the given mass, and $$M$$ is the molar mass.

Using the atomic mass of chlorine, $$M_{\text{Cl}} = 35.5\ \text{g mol}^{-1},$$ we have

$$n_{\text{Cl}} \;=\;\frac{0.0355\ \text{g}}{35.5\ \text{g mol}^{-1}}.$$

Carrying out the division,

$$n_{\text{Cl}} = 0.001\ \text{mol}.$$

Each mole contains Avogadro’s number of entities. We therefore multiply the moles of chlorine by Avogadro’s constant to find the number of chlorine atoms.

Formula: $$N = n \times N_{\!A},$$ where $$N$$ is the number of entities and $$N_{\!A} = 6.023 \times 10^{23}\ \text{mol}^{-1}.$$

Substituting the values,

$$N_{\text{Cl atoms}} \;=\;0.001\ \text{mol}\times 6.023 \times 10^{23}\ \text{mol}^{-1}.$$

Multiplying,

$$N_{\text{Cl atoms}} = 6.023 \times 10^{20}.$$

Therefore, 1 g of the chlorohydrocarbon contains $$6.023 \times 10^{20}$$ chlorine atoms.

Hence, the correct answer is Option D.