Question

Enter your answer in the provided box. Calculate the vapor pressure of a solution of 43.5 g of glycerol (C3H803) in 500.0 g of water at 25°C. The vapor pressure of water at 25°C is 23.76 torr. (Assume ideal behavior.) I torr

Answer 1

Concept: The solution behaves ideally, so glycerol can be assumed as non-volatile solue and Raoult's law for non-volatile solutes can be applied.

Note that Raoult's law for non-volatile solute states that relative lowering of vapour pres­sure of a solution containing a non­volatile solute is equal to the mole fraction of the solute in the solution

That is :

       ......where, P* = vapor pressure of pure solution

P - P. - = Xsolute

                                                             P = vapor pressure of solution

Calculations:

Given: P* = 23.76 torr (for pure water)

mass of glycerol = 43.50 g moles of glycerol = (43.50 g ) / (92.09 g/mol) = 0.4724           .. as molar mass of glcerol = 92.09 g/mol

mass of water = 500 g moles of water = (500.0g ) / (18.01 g/mol)   = 27.76          .. as molar mass of water = 18.01 g/mol

So. mole fraction of glycerol:

0.4724 1794, 92 70 = 0.01673 Xglycerol =

and

Xwater = 1-0.01673 = 0.9833

From Raoult's law expression above:

PS D# = Xglycerol Di =1-Xglycerol = Xwater

So,

Ps = Xwater x P* = 0.9833 x 23.76torr = 23.36torr

Rounding to 4 significant digits:

Ps = 23.36 Torr    ....Answer

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