0.078 g of a hydrocarbon occupy 22.414 mL of volume at S.T.P. The empirical formula of the hydrocarbon is CH. The molecular formula of hydrocarbon is

Options

A. `C_(2)H_(2)`
B. `C_(6)H_(6)`
C. `C_(8)H_(8)`
D. `C_(4)H_(4)`

Answer

B

Explanation

22.414 mL of gaseous hydrocarbon at S.T.P. has mass = 0.078 g
22414 mL of gaseous hydrocarbon at S.T.P. will have mass = 78 g
`:.` Molar mass of hydrocarbon `= 78 gmol^(-1)`
Empirical formula mass of hydrocarbon (CH)
`= 12+1=13`
`:. n = ("Molecular mass")/("E.F. mass") = (78)/(13) =6`
`:.` Molecular mass of hydrocarbon
`= (CH)_(6)=C_(6)H_(6)`