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Gennadij [26K]
3 years ago
15

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

Chemistry
1 answer:
Artist 52 [7]3 years ago
8 0

<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)}}

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

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