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

Answer

C

Explanation

% of chlorine = 3.55
Thus, in 100 g of chlorohydrocarbon, mass of chlorine = 3.55 g
1 g of chlorohydrocarbon will contain `= (3.55)/(100)g` of chlorine
`:.` No. of moles of chlorine atoms `= ("Mass")/("Atomic mass") = (3.55)/((100)/(35.5)) = 1 xx 10^(-3)`
1 mole of chlorine contains `6.023 xx 10^(23)` chlorine atoms. `:. 1 xx 10^(-3)` mole of chlorine will contain
`1 xx 10^(-3) xx 6.023 xx 10^(23) = 6.023 xx 10^(20)` chlorine atoms