Question

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

• A. $6.023\times 10^{20}$
• B. $6.023\times 10^9$
• C. $6.023\times 10^{21}$
• D. $6.023\times 10^{23}$

Answer 1

Answer: (A)

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

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

Number of moles of chlorine can be calculated as

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

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