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} $
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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} $
