<u>Answer:</u> The molarity of solution is 0.274 M and the osmotic pressure of the solution is 6.70 atm
<u>Explanation:</u>
To calculate the molarity of the solution, we use the equation:
![\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20the%20solution%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20solute%7D%5Ctimes%201000%7D%7B%5Ctext%7BMolar%20mass%20of%20solute%7D%5Ctimes%20%5Ctext%7BVolume%20of%20solution%20%28in%20mL%29%7D%7D)
We are given:
Given mass of estrogen = 13.5 g
Molar mass of estrogen = 272.40 g/mol
Volume of solution = 181 mL
Putting values in above equation, we get:
![\text{Molarity of solution}=\frac{13.5\times 1000}{272.40\times 181}\\\\\text{Molarity of solution}=0.274M](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20solution%7D%3D%5Cfrac%7B13.5%5Ctimes%201000%7D%7B272.40%5Ctimes%20181%7D%5C%5C%5C%5C%5Ctext%7BMolarity%20of%20solution%7D%3D0.274M)
Hence, the molarity of solution is 0.274 M
To calculate the osmotic pressure of the solution, we use the equation:
![\pi=iMRT](https://tex.z-dn.net/?f=%5Cpi%3DiMRT)
where,
= osmotic pressure of the solution = ?
i = Van't hoff factor = 1 (for non-electrolytes)
M = molarity of solute = 0.274 M
R = Gas constant = ![0.0821\text{ L atm }mol^{-1}K^{-1}](https://tex.z-dn.net/?f=0.0821%5Ctext%7B%20L%20atm%20%7Dmol%5E%7B-1%7DK%5E%7B-1%7D)
T = temperature of the solution = 298 K
Putting values in above equation, we get:
![\pi=1\times 0.274mol/L\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\\pi=6.70atm](https://tex.z-dn.net/?f=%5Cpi%3D1%5Ctimes%200.274mol%2FL%5Ctimes%200.0821%5Ctext%7B%20L.atm%20%7Dmol%5E%7B-1%7DK%5E%7B-1%7D%5Ctimes%20298K%5C%5C%5C%5C%5Cpi%3D6.70atm)
Hence, the osmotic pressure of the solution is 6.70 atm