The empirical formula of a non-electrolyte is `CH_(2)O`. A solution containing 3 g `L^(-1)` of the compound exerts the same osmotic pressure as that of 0.05 M glucose solution. The molecular formula of the compound is :

Options

A. `CH_(2)O`
B. `C_(2)H_(4)O_(2)`
C. `C_(4)H_(8)O_(4)`
D. `C_(3)H_(6)O_(3)`

Answer

B

Explanation

Concentration of compouinds in solution `= 3g L^(-) = (3)/(M) mol L^(-1)`
Since it is isotonic with `0.05M` glucose solution `(3)/(M) = 0.05` or `M = 60`
Empirical formula mass of `CH_(2)O = 30`
`n = ("Mol .mass")/(EF"mass") = (60)/(30) = 2`
`:.` molecular formula `= 2 xx CH_(2)O = C_(2)H_(4)O_(2)`