Answer: 11.6g of estrogen are needed to generate an osmotic pressure of 4.45 atm when dissolved in 234 ml of a benzene solution at 298 K.
Explanation:
To calculate the amount of solute, we use the equation for osmotic pressure, which is:
[tex]\pi=iMRT[/tex]
Or,
[tex]\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT[/tex]
where,
[tex]\pi[/tex] = osmotic pressure of the solution = 4.45 atm
i = Van't hoff factor = 1 (for non-electrolytes)
Let Mass of solute (estrogen) = x g
Volume of solution = 234 mL
R = Gas constant = [tex]0.0821Latmmol^{-1}K^{-1}[/tex]
T = temperature of the solution = [tex]298K[/tex]
Putting values in above equation, we get:
[tex]4.45=1\times \frac{x\times 1000}{272.4\times234}\times 0.0821Latmmol^{-1}K^{-1}\times 298K[/tex]
[tex]x=11.6g[/tex]
Hence, 11.6g of estrogen are needed to generate an osmotic pressure of 4.45 atm when dissolved in 234 ml of a benzene solution at 298 K.
