At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bar at the same temperature, what would be its concentration? [Report your answer upto two decimal places.]

Step-by-step solution

π = CRT = W B × R × T M B × V For both solutions, R, T and V are constants. For I solution 4.98 bar = 36g × R × T 180 g mol - 1 × V ...(i) For II solution 1.52 bar = W B × R × T M B × V ...(ii) On dividing Eq. (ii) by Eq. (i), we get 1.52 bar 4.98 bar = W B × R × T M B × V × 180 × V 36 × R × T W B M B = 1.52 4.98 × 5 = 0.0610 mol L - 1