Answer to Question #115829 in General Chemistry for Jonathan

Question

3.) 3.8 mols of Kool-Aid are added to a jug containing 3.28 liters of H2O at a temperature of 10°C. The vapor pressure of water at 10°C is 9.2 mmHg. What is the new vapor pressure of the Kool-Aid solution?

Answer 1

P_pure H2O = 9.2 mm Hg

To solve for the mole fraction, you must first convert the 3.28 L of water into moles.  

1 L = 1000 mL = 1000 g Knowing this, you can convert the mass of water (3280 g) into moles:

3280 g / 18.02 g (molar mass of water) = 182.0 moles H₂O

Solve for the mole fraction, x_H2O.

x_H2O = moles H₂O / total moles = 182.0 moles / 182.0 + 3.8 moles = 0.979 Finally, apply Raoult's Law.

P_Kool-Aid = x_H2O P_pure H2O = (0.979) x (9.2 mm Hg) = 9.0 mm Hg