The vapor pressure of water at 20°C is 17.54 torr. What is the vapor pressure (in torr)of a solution containing 143.8g of dextrose (C6H12O6 ) and 146.5 g of water at ?

Question

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Answer 1

Let's assume the vapor pressure of water after the addition of dextrose is still to be determined at 20 °C (constant temperature).

We'll need Raoult's Law: states that the partial pressure of each component of an ideal mixture of liquids is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture.

So, the partial pressure of water in the mixture is equal to the vapor pressure of pure water multiplied by its mole fraction in the mixture.

P_{water in mixture} = X_{water in mixture} * P_{pure water}

You are given P_{pure water}

and you need to solve for X_{water in mixture} in order to determine the vapor pressure of water in the mixture.

X_{water in mixture} is the mole fraction of water in the mixture

So, you'll need the moles of 146.5 g of water AND the moles of 143.8 g of dextrose (C₆H₁₂O₆)

Determine the moles of each using their corresponding molar masses

Once you have the moles ...

For example, if you have 5.0 moles of water and 10.0 moles of dextrose,

the mole fraction of water in the mixture can be determine by

moles of water over total moles (water + dextrose) ... 5.0 moles / 15.0 moles = 1/3 = 0.33

notice the "moles" in the numerator divided by the "moles" in the denominator equal "1" and 0.33 has no units

With the mole fraction and the given pressure of pure water, you can determine the vapor pressure of water in the mixture