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Calculate the vapor pressure of a solution of 32.5 g of glycerol (C3H8O3) in 500.0 g of water at 25°C. The vapor pressure of water at 25°C is 23.76 torr. (Assume ideal behavior.)

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The vapor pressure is obtained as 23.47 torr.

What is the vapor pressure?

Given that; p = x1p°

p = vapor pressure of the solution

x1 = mole fraction of the solvent

p° = vapor pressure of the pure solvent

Δp = p°(1 - x1)

Δp =x2p°

Δp =  vapor pressure lowering

x2 = mole fraction of the  of the solute

Number of moles of  glycerol =  32.5 g/92 g/mol = 0.35 moles

Number of moles  of water = 500.0 g/18 g/mol = 27.8 moles

Total number of moles = 0.35 moles + 27.8 moles = 28.15 moles

Mole fraction of glycerol = 0.35 moles/28.15 moles = 0.012

Mole fraction of water = 27.8 moles/28.15 moles =0.99

Δp =  0.012 * 23.76 torr

Δp =  0.285 torr

p1 = p° - Δp

p1 = 23.76 torr -  0.285 torr

p1 = 23.47 torr

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