### JEE Main On 16 April 2018 Question 6

##### 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}} $ )[JEE Main 16-4-2018]

#### Options:

A) $ 6.023\times 10^{23} $

B) $ 6.023\times 10^{21} $

C) $ 6.023\times 10^{9} $

D) $ 6.023\times 10^{20} $

## Show Answer

#### Answer:

Correct Answer: D

#### Solution:

An unknown chlorohydrocarbon has 3.55% of chlorine.

100 g of chlorohydrocarbon has 3.55 g of chlorine.

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=0.001,mol\times 6.023\times 10^{23}mo{l^{-1}}=6.023\times 10^{20} $