Osmotic pressure of 40% (wt./vol.) urea solution is 1.64 atm and that ...
Solution:
Given,
Osmotic pressure of 40% (wt./vol.) urea solution = 1.64 atm
Osmotic pressure of 3.42% (wt./vol.) cane sugar solution = 2.46 atm
When equal volumes of the above two solutions are mixed, the osmotic pressure of the resulting solution is to be calculated.
Let us assume the volume of each solution to be V.
Then, the number of moles of urea present in V volume of 40% (wt./vol.) urea solution = 40/60.02 × V × 1000/60 = 0.666 V
Similarly, the number of moles of cane sugar present in V volume of 3.42% (wt./vol.) cane sugar solution = 3.42/342.3 × V × 1000/180 = 0.019 V
On mixing, the total volume of the solution = 2V
The total number of moles of solute in the mixture = 0.666 V + 0.019 V = 0.685 V
The concentration of the resulting solution = (0.685V/2V) × 100 = 34.25%
Using the formula, π = CRT, where π is the osmotic pressure, C is the concentration, R is the gas constant and T is the temperature, we can find the osmotic pressure of the resulting solution.
Let us assume the temperature to be constant and equal to room temperature, which is approximately 25°C or 298 K.
Therefore,
For the urea solution, π1 = 1.64 atm
For the cane sugar solution, π2 = 2.46 atm
For the resulting solution, C = 34.25%
R = 0.0821 L atm mol^-1 K^-1 (gas constant)
T = 298 K
π3 = CRT = (34.25/100) × 0.0821 × 298 = 2.05 atm
Therefore, the osmotic pressure of the resulting solution is 2.05 atm, which is option D.