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What is the osmotic pressure of the solution?

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18.6 grams of solute with molecular mass of 8940 grams are dissolved in enough water to make 1.00dm^3 of solution at 25 c. TT ______ atmospheres

Oct 23rd, 2014

TT = iMRT, where: TT ------> osmotic pressure in atm.

                                i ---------> Van't Hoff factor for the solute (=1 if it does not disassociate)

                               M ---------> concentration in mol/l

                               R ---------> universal gas constant  = 0.082   litres.atm/mol.K

                               T ---------> absolute temperature in kelvin

1) Determine the molecular mass of the compound ------> the sum of (the number of each type of atom* its atomic mass )

in this case it is given by 8940 g. 

Technically, the molecular mass is the molar mass expressed in g/mol

So, molar mass  = 8940 g/mol.

2) Calculate the number of moles (n) using the given mass of the solute :          8940 g --------> 1 mol

                                                                                                                                 18.6 g ---------> ?? mol

 therefore n = 18.6 * 1/8940 =0.0021 mol

3) Calculate the concentration M:   M = n / V ,  where V in litres  ( 1dm^3 = 1L)

                                                            = 0.0021 / 1 = 0.0021 mol/L

4) Calculate the absolute temperature (T) in K :

   T (K) = t (c) +273 ,    so   T = 25 + 273 = 298 K

5) Assume i=1 

6) Substitute in the main formula to calculate TT

TT = 1 * 0.0021 * 0.082 * 298 = 0.513 atm

                                                                         

Oct 23rd, 2014

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