Question

An aqueous solution contain 1.00g/L of a derivative of the detergent lauryl alcohol. The osmotic pressure of this solution at 25°C is 2.0×10^3. What is the molar mass of the detergent (R = 8.314jmol-1k-1

Answer 1

M = moles 
R = 0.0821 L*atm/mol*K 
T = temperature (in Kelvin) 

So since R is in atm and Kelvin, you must convert. 
25°C + 273 = 298K 
17.8 torr / 760 = .0234atm 
and Moles = Mass/Molar Mass 
(assume mass of 1g) 

Now plug it in: 
.0234atm = (1g/Molar Mass) x .0821 L*atm/mol*K x 298K 

Rearrange and solve for Molar Mass: 
1044 g/mole