Question

Osmotic Pressure Of 30% Solution Of Glucose Is 1.20 Atm And That Of 3.42%Solution Of Cane Sugar Is 2.5Atm . The Osmotic Pressure Of The Mixture Containing Equal Volume Of The Two Solution Will Be

Answer 1

Answer:

1.85

Explanation:

Let M1 be the molarity of glucose and

LEt M2 be the molarity for Cane -sugar solution

When equal volumes would be mixed then,  -

M1V + M2V = M(2V)

Thus, the new concentration will be -

M = (M1+M2)/2

 From the first solution

1.20 =M1ST

Similarly,  

M2 = 2.50/ST

Substituting it in M and applying for the mixed solution.

O.P = 1*M*S*T , where S*T will be cancelled out

Thus O.P = 3.7/2(ST)

ST = 3.7/2

ST = 1.85 atm