Question

An unknown chlorohydrocarbon has $3.55\%$ of chlorine. If each molecule of the hydrocarbon has one chlorine atom only, then chlorine atoms present in $1g$ of chlorohydrocarbon are:

[Atomic wt. of $Cl=35.5u$; Avogadro constant $=6.023\times {10}^{23}{mol}^{-1}$]

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

Answer 1

The correct option is D $6.023\times {10}^{20}$

An unknown chlorohydrocarbon has $3.55$% of chlorine.

100 g of chlorohydrocarbon has $3.55$ g of chlorine.

1 g of chlorohydrocarbon will have $3.55\times \frac{1}{100}=0.0355$ g of chlorine.

Atomic wt. of $Cl=35.5$ g/mol

Number of moles of $Cl=\frac{0.0355g}{35.5g/mol}=0.001$mol

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

Hence, option $D$ is correct.