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

Step-by-step explanation:

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

`\text{Molarity of the solution}=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \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}=(13.5* 1000)/(272.40* 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* 0.274mol/L* 0.0821\text{ L.atm }mol^(-1)K^(-1)* 298K\\\\\pi=6.70atm`

Hence, the osmotic pressure of the solution is 6.70 atm