osmotic pressure of 30% solution of glucose is 1.20 atm and that of 3.42 % of cane sugar is 2.5 atm. The osmotic pressure of the solution containing the equal volume of both the solution will be?

let M1 be the molarity of glucose. and M2 be the molarity for Cane -sugar . respectively. Of equal volumes would be mixed then, M1V + M2V = M(2V) M = (M1+M2)/2 would be the new concentration. From first soln. 1.20=M1ST or M1 = 1.20/ST similarly M2 = 2.5/ST Put it in M. then apply for mixed soln. O.P = 1*M*S*T S*T would be canceled out. O.P=3.7/(2*ST) *ST=3.7/2=1.85atm

Since volume is same therefore , M= M1+M2/2

Osmotic pressure of 30% solution of glucose is 1.20 ATM and death of 3.42 percent solution of Cane sugar is 2.5 ATM the Osmotic pressure of the mixture containing equal volumes of two solution will be