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. `C_(2)H_(4)O_(2)`
B. `C_(3)H_(6)O_(3)`
C. `C_(5)H_(10)O_(5)`
D. `C_(4)H_(8)O_(4)`

Answer

D

Explanation

For isotonic solution : `(w_(1))/(m_(1)V_(1))=(w_(2))/(m_(2)V_(2))`
`w_(1)` = mass of glucose = `0.05xx180g=9g`
`m_(1)` = molecular mass of glucose = 180 g
Assuming `V_(1)=V_(2)=1L`
`w_(2)` = mass of compound = 6g
`m_(2)` = molecular mass of compound = ?
`therefore (9)/(180)=(6)/(x)impliesx=120g`
`therefore ("Molecular mass")/("Empirical mass")=nimpliesn=(120)/(30)=4`
`therefore` Molecular formula = `C_(4)H_(8)O_(4)`.