In equimolar solution of glucose, NaCl and `BaCl_(2)` , the order of osmotic pressure at same temperature is as follow `:`

Options

A. Glucose `gt` NaCl `gt BaCl_(2)`
B. NaCl `gt BaCl_(2) gt` Glucose
C. `BaCl_(2) gt` NaCl `gt` Glucose
D. Glucose `gt BaCl_(2) gt` NaCl

Answer

C

Explanation

`BaCl_(2)=Ba^(2+)+2Cl^(-)=3` ion
`NaClimpliesNa^(+)+Cl^(-)=2` ion
Glucose implies No ionisation
`therefore BaCl_(2)gtNaClgt` Glucose