If 3 ~g of glucose (molar mass =180 ~g ) is dissolved in 60 ~g of water at 15^ C , the osmotic pressure of the solution will be

Step-by-step solution

Given, molar mass of glucose M_B=180 ~g Mass of glucose, W_B=3 ~g Mass of water, W_A=60 ~g Temperature =15^ C 273+15=288 ~K Osmotic pressure, = ? C= W_B 1000 W_A M_B = 3 1000 60 180 =0.277 ~mol ~L ⁻¹ We know that, aligned & =C R T=0.277 0.0821 288 & =6.549 6.57 ~atm aligned