Colligative Properties of Nonelectrolyte Solutions

[erik-the-red]

How many grams of sucrose ($$C_{12} H_{22} O_{11}$$) must be added to 552 g of water to give a solution with a vapor pressure of 2.0 mmHg less than that of pure water at $$20 ^ \circ$$ C? The vapor pressure of pure water at $$20 ^ \circ$$ C is 17.5 mmHg.

This is one of many similar types of problems in Chemistry (7th ed.) by Raymond Chang.

I don't know how to start this. Does it involve Raoult's Law?

 

[Gokul43201]

Yes, (in the limit of a dilute solution) you can use Raoult's Law.

PS : Next time, such questions go in the appropriate section of the Homework/Coursework forum. And before posting there, please read the rules for posting (also found in the second line of my sig.)

 

[Blueshift5]

Yes, this problem does involve Raoult's Law, which is a colligative property that describes the relationship between the vapor pressure of a solution and the mole fraction of the solute present. In this case, we are looking for the amount of sucrose (a nonelectrolyte solute) needed to decrease the vapor pressure of water by 2.0 mmHg.

To solve this problem, we can use the following equation:

ΔP = Xsolute * P°solute

Where ΔP is the change in vapor pressure, Xsolute is the mole fraction of the solute, and P°solute is the vapor pressure of the pure solute.

First, we need to calculate the mole fraction of sucrose in the solution. This can be done by dividing the moles of sucrose by the total moles of both water and sucrose.

Moles of sucrose = mass of sucrose / molar mass of sucrose
= 552 g / 342.3 g/mol
= 1.61 mol

Total moles of solution = moles of sucrose + moles of water
= 1.61 mol + 552 g / 18.02 g/mol
= 30.59 mol

Mole fraction of sucrose = 1.61 mol / 30.59 mol
= 0.0527

Next, we can plug in the values into the equation for Raoult's Law:

ΔP = 0.0527 * 17.5 mmHg
= 0.921 mmHg

Since we want the vapor pressure to decrease by 2.0 mmHg, we can set up a proportion to find the mass of sucrose needed:

0.921 mmHg / 1.61 mol = 2.0 mmHg / x mol

Solving for x, we get:

x = 1.61 * 2.0 mmHg / 0.921 mmHg
= 3.49 mol

Finally, we can convert the moles of sucrose back to grams:

Mass of sucrose = moles of sucrose * molar mass of sucrose
= 3.49 mol * 342.3 g/mol
= 1193 g

Therefore, 1193 grams of sucrose must be added to 552 grams of water to give