A camphor sample melts at `176^(@)C.K_(f)` for camphor is 40 K kg `mol^(-1)`. A solution of 0.02 g of a hydrocarbon in 0.8 g of camphor melts at `156.77^(@)C`. The hydrocarbon is made up of 92.3% of carbon. What is the molecular formula of the hydrocarbon?

Options

A. `C_(6)H_(6)`
B. `C_(12)H_(12)`
C. `C_(4)C_(4)`
D. `C_(8)H_(8)`

Answer

C

Explanation

Melting point of camphor ` = 176^(@)` C
` (K_(f))` for camphor = 40 k kg mol `""^(-1)`
Mass of hydrocarbon ` (w_(B)) = 0.02 g `
Mass of camphor ` (W_(A)) = 0.8 g `
Thus , ` Delta T_(f)` for camphor ` = 176 - 156.77 `
` = 19.23^(@)C = 19.23 K `
` therefore ` Depression in freezing point ,
`DeltaT_(f) = (K_(f) xx W_(B))/(M_(B)) xx (1000)/(W_(A))`
`19.23 = (40 xx 0.02 xx 1000)/(M_(g) xx 0.8)`
Molar mass of solute ` (M_(g))`
` = (40 xx 0.02 xx 1000)/(19.23 xx 0.8) = 52 `

`therefore ` Empirical formula = CH
Empirical formula weight = 12 + 1 = 13
Multiple `(n) = ("Molecular weight")/("Empirical formula weight") = (52)/(13) = 4 `
Thus, molecular formula `= (CH)_(4) = C_(4) H_(4)` .