Question

An unknown chlorohydrocarbon as $3.55$% 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$=6.023\times {10}^{23}{mol}^{-1}$)

• A. $6.032\times {10}^9$
• B. $6.032\times {10}^{23}$
• C. $6.032\times {10}^{21}$
• D. $6.032\times {10}^{20}$

Answer 1

Answer: (D)

$6.032\times {10}^{20}$

An unknown chlorohydrocarbon has 3.55% of chlorine.
100 g of chlorohydrocarbon has 3.55 g of chlorine.
1 g of chlorohydrocarbon will have $3.55\times \frac{1}{100}=0.0355g$
Atomic wt. of $Cl$ = 35.5g/mol
Number of moles of $Cl=\frac{0.0355g}{3.55g/mol}=0.001mol$
Number of atoms of $Cl$ = $0.001mol\times 6.023\times {10}^{23}mo{l}^{-1}=6.023\times {10}^{20}$