An unknown chlorohydrocarbon has 3.55% of chlorine. If each molecule of the hydrocarbon has one chlorine atom only, chlorine atoms present in 1 g of chlorohydrocarbon are :
(Atomic wt. of Cl = 35.5 u,
Avogardo constant `=6.023xx10^(23) "mol"^(-1)`)

Options

A. `6.023 xx 10^(9)`
B. `6.023 xx 10^(23)`
C. `6.023 xx 10^(21)`
D. `6.023 xx 10^(20)`

Answer

D

Explanation

Given percentage of chlorine in an hydrocarbon = 3.55% i.e.,
100 g of chlorohydrocarbon has 3.55 g of chlorine.
1 gof chlorohydrocarbon will have `(355)/(100)= 0.0355 g` of chlorine.
Atomic wt. of `CI=35.5 g // mol`
Number of moles of `Cl =(0.0355 g)/(35.5 g // mol) = 0.001` mole
Number of atons of `CI=0.001" mole" xx 6.023 xx 10^(23) mol^(-1)`
`= 6.023 xx 10^(20)`