Question

The nonvolatile, nonelectrolyte estrogen (estradiol), C18H24O2 (272.40 g/mol), is soluble in chloroform CHCl3.

Calculate the osmotic pressure generated when 13.5 grams of estrogen are dissolved in 181 ml of a chloroform solution at 298 K.

The molarity of the solution is ? M.
The osmotic pressure of the solution is ? atmospheres.

Answer 1

Answer: The molarity of solution is 0.274 M and the osmotic pressure of the solution is 6.70 atm

Explanation:

To calculate the molarity of the solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

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:

$\text{Molarity of solution}=\frac{13.5\times 1000}{272.40\times 181}\\\\\text{Molarity of solution}=0.274M$

Hence, the molarity of solution is 0.274 M

To calculate the osmotic pressure of the solution, we use the equation:

$\pi=iMRT$

where,

$\pi$ = osmotic pressure of the solution = ?

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = 0.274 M

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = 298 K

Putting values in above equation, we get:

$\pi=1\times 0.274mol/L\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\\pi=6.70atm$

Hence, the osmotic pressure of the solution is 6.70 atm