At 27 ° C , a solution containing 2 . 5 g of solute in 250 . 0 mL of solution exerts an osmotic pressure of 400 Pa . The molar mass of the solute is g mol - 1 (Nearest integer) (Given : R = 0 . 083 L bar - 1 mol - 1 )

Step-by-step solution

The osmotic pressure of a non-electrolytic solution can be calculated as follows, π = CRT π   = osmotic pressure C = molarity 400   Pa 10 5 = 2 . 5   g M o 250 / 1000   L × 0 . 83 L - bar K . mol × 300   K ⇒   400   ×   10 - 5 ( 1 . 01325 )   =   2 . 5   ×   1000 M o   ×   250   0 . 083   ×   300 ⇒   400   ×   10 - 5   =   10 M o   ×