Question

The nonvolatile, nonelectrolyte testosterone , C₁₉H₂₈O₂ (288.4 g/mol), is soluble in chloroform, CHCl₃.

Calculate the osmotic pressure (in atm) generated when 14.5 grams of testosterone are dissolved in 188 mL of a chloroform solution at 298 K.

In a laboratory experiment, students synthesized a new compound and found that when 12.44 grams of the compound were dissolved to make 217.2 mL of a diethyl ether solution, the osmotic pressure generated was 1.57 atm at 298 K. The compound was also found to be nonvolatile and a nonelectrolyte.


What is the molecular weight they determined for this compound?

Answer 1

Mass of testosterone = 14.5 g

Molar mass of testosterone = 288.4 g/mol

Moles of testosterone = mass /molar mass = 14.5 g/288.4 g/mol = 0.0503 moles

Volume = 188mL = 0.188L

Molarity (M) = Moles/Volume = 0.0503moles/0.188L = 0.267 M

Since testosterone is non-electrolyte. So, i = 1

Pressure () = iMRT

= 1* 0.267 M*(0.08206 atm.L/mol.K)*298K = 6.54 atm

2. = iMRT

1.57atm = 1* M*(0.08206 atm.L/mol.K)*298K

M = 0.0642 mol/L

But M = mass/(Molar mass *Volume)

0.0642mol/L = 12.44g/(Molar mass *0.2172L)

Molar mass = 12.44g(0.0642*0.2172) = 892.1 g/mol