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 = \(6.023 \times 10^{23}\ \mathrm{mol^{-1}}\))
(Atomic wt. of Cl = 35.5 u; Avogadro constant = \(6.023 \times 10^{23}\ \mathrm{mol^{-1}}\))
Show Hint
- \(6.023 \times 10^{20}\)
- \(6.023 \times 10^{9}\)
- \(6.023 \times 10^{21}\)
- \(6.023 \times 10^{23}\)
The Correct Option is A
Solution and Explanation
Moles of Cl atoms = \(\frac{0.0355}{35.5} = 10^{-3}\) mol.
Number of Cl atoms = \(10^{-3}\times 6.023\times 10^{23} = 6.023\times 10^{20}\).
Answer: (1) \(6.023 \times 10^{20}\)
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