If `0.765 g` of an acid gives `0.535 g` of `CO_(2)` and `0.138 g` of `H_(2)O`, then the ratio of the percentage of `C` to `H` is

Options

A. `19:2`
B. `18:11`
C. `20:17`
D. `1:7`

Answer

A

Explanation

`% C = (12)/(44) xx (m_(CO_(2)))/(m_(o .c.)) xx 100%`
`= (12)/(44) xx (0.535 g )/(0.765 g) xx 100% = 19.07%`
`% H = (2)/(18) xx (m_(H_(2)O))/(m_(o .c.)) xx 100%`
`= (2)/(18) xx (0.138 g)/(0.765 g) xx 100%`
`= 2%`
Thus, the ratio of `%` of `C` to `H` is `19:2` (approx).