Osmotic pressure of 30% solution of glucose is 1.20 atm and that of 3.42% solution of cane sugar is 2.5 atm. The osmotic pressure of the mixture containing equal volumes of the two solutions will be

Question

Osmotic pressure of 30% solution of glucose is 1.20 atm and that of 3.42% solution of cane sugar is 2.5 atm. The osmotic pressure of the mixture containing equal volumes of the two solutions will be (A) 2.5 atm (B) 3.7 atm (C) 1.85 atm (D) 1.3 atm

Tardigrade Admin · Accepted Answer

πf Vf=π1 V1+π2 V2 πf =(1.2V+2.5V/2V) =(3.7V/2V) πf=1.85 atm

Q. Osmotic pressure of 30% solution of glucose is 1.20 atm and that of 3.42% solution of cane sugar is 2.5 atm. The osmotic pressure of the mixture containing equal volumes of the two solutions will be

2.5 atm

3.7 atm

1.85 atm

1.3 atm

$π_{f} V_{f} = π_{1} V_{1} + π_{2} V_{2}$
$π_{f} = \frac{1.2 V + 2.5 V}{2 V}$
$= \frac{3.7 V}{2 V}$
$π_{f} = 1.85$ atm

Q. Osmotic pressure of 30% solution of glucose is 1.20 atm and that of 3.42% solution of cane sugar is 2.5 atm. The osmotic pressure of the mixture containing equal volumes of the two solutions will be

2.5 atm

3.7 atm

1.85 atm

1.3 atm

πf​Vf​=π1​V1​+π2​V2​ πf​=2V1.2V+2.5V​ =2V3.7V​ πf​=1.85 atm

$π_{f} V_{f} = π_{1} V_{1} + π_{2} V_{2}$
$π_{f} = \frac{1.2 V + 2.5 V}{2 V}$
$= \frac{3.7 V}{2 V}$
$π_{f} = 1.85$ atm