Question

The nonvolatile, nonelectrolyte testosterone, C19H28O2 (288.40 g/mol), is soluble in benzene C6H6. Calculate the osmotic pressure generated when 12.4 grams of testosterone are dissolved in 168 ml of a benzene solution at 298 K.

Answer 1

Answer: 6.26atm

Explanation:Please see attachment for explanation

Answer 2

Answer:

The osmotic pressure is 6.26 atm

Explanation:

Step 1: Data given

Mass of testosterone = 12.4 grams

Volume of benzene = 168 mL

Temperature = 298 Kelvin

Step 2: Calculate moles of testosterone

Moles testosterone = mass / molar mass

Moles testosterone = 12.4 grams / 288.42 g/mol

Moles testosterone = 0.0430 moles

Step 3: Calculate molarity

Molarity = moles / volume

Molarity = 0.0430 moles / 0.168 L

Molarity = 0.256 M

Step 4 : Calculate the osmotic pressure

π = iMRT

⇒ with i = The Van't hoff factor = 1 (since testosterone is nonelectrolyte)

⇒ with M = the molair concentration = 0.256 M

⇒ with R = the gas constant = 0.08206 L*atm/k*mol

⇒ with T = the temperature = 298 K

π = 1*0.256*0.08206*298

π = 6.26 atm

The osmotic pressure is 6.26 atm