The nonvolatile, nonelectrolyte testosterone, C19H28O2 (288.40 g/mol), is soluble in benzene C6H6. Calculate the osmotic pressure generated when 12.9 grams of testosterone are dissolved in 286 ml of a benzene solution at 298 K.

3.824 atm

Explanation:

From the ideal gas equation

P = mRT/MW × V

m is mass of testosterone = 12.9 g

R is gas constant = 82.057 cm^3.atm/mol.K

T is temperature of benzene solution = 298 K

MW is molecular weight of testosterone = 288.40 g/mol

V is volume of benzene solution = 286 ml = 286 cm^3

P = 12.9×82.057×298/288.4×286 = 3.824 atm

? you expect us to answer this? ? this is your schoolwork, i don't think we can with this, i think it would considering being

i just learnt this recently but i need to look through my notes again