Question

If \(X\) is a haloalkane with a single chlorine atom per molecule and the percentage of Cl is 55, what would be the number of Cl atoms present in 0.1 g of the haloalkane? Atomic mass of Cl = 35.5 g/mol

• A. \(6.022 \times 10^{22}\)
• B. \(1.2044 \times 10^{21}\)
• C. \(9.328 \times 10^{20}\)
• D. \(9.329 \times 10^{23}\)

Hint:

When only one atom of an element is present per molecule, number of atoms equals number of molecules. Always use Avogadro’s number for conversion.

Answer 1

The Correct Option is C

Step 1: Use percentage composition.
Percentage of chlorine is given as:
\[ 55\% \]
So, in 100 g of compound, chlorine present is:
\[ 55\ \text{g} \]

Step 2: Find molar mass of compound.
Since there is only one Cl atom per molecule:
\[ \text{Molar mass of compound} = \frac{35.5 \times 100}{55} \]
\[ = 64.54\ \text{g/mol (approx.)} \]

Step 3: Find moles of compound in 0.1 g.
\[ \text{moles} = \frac{0.1}{64.54} \]
\[ \approx 1.55 \times 10^{-3}\ \text{mol} \]

Step 4: Use Avogadro's number.
Number of molecules:
\[ = 1.55 \times 10^{-3} \times 6.022 \times 10^{23} \]
\[ \approx 9.328 \times 10^{20} \]

Step 5: Relate molecules to chlorine atoms.
Each molecule contains one chlorine atom.
So, number of Cl atoms = number of molecules.

Step 6: Final calculation.
\[ \text{Number of Cl atoms} = 9.328 \times 10^{20} \]

Step 7: Final Answer.
\[ \boxed{9.328 \times 10^{20}} \]