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

Conc. Of compound in solution = `3 gL^(-1)`
`=(3)/(M)"mol L"^(-1)`
At it is isotonic with 0.05 M glucose solution,
`(3)/(M)=0.05 or M = 60`
Empirical formula mass of `CH_(2)O=30`
`therefore n=("Mol. Mass")/("E.F. mass")=(60)/(70)=2`
Hence, molecular formula `=2xxCH_(2)O=C_(2)H_(4)O_(2)`