Elevation in boiling point of an aqueous glucose solution is 0.6, `K_b` for water is 0.52 K molal`""^(-1).` The mole fraction of glucose in the solution is:

Options

A. `0.02`
B. `0.03`
C. `0.01`
D. `0.04`

Answer

A

Explanation

`Delta T= K_(b) xx M`
`M= (0.6)/(0.52)= 1.154`
1.154 moles in 1000g i.e, 55.56 moles water.
Mole fraction `= (1.154)/(56.7) ~~ 0.02`