Question

Please explain, Im so confused!

A. What is the vapor pressure in atm of a solution at 25 oC produced by dissolving 138.4 g of dextrose (C6H12O6) in 376.4 g of water? Vapor pressure of water at 25 oC = 0.0313 atm

B. An 18.3 g of an unknown non-ionizing solute is added to 88.8 g of water. If the boiling point of the solution is 101.4, what is the molar mass of the solute?

C. How many moles of ions are present in an ideal solution that is produced by dissolving 18.1 g of Cu(NO₃)₂ in 341 g of water?

Answer 1

a). Moles dextrose = 75.2 gms over 180 grams per mole = 0.769 moles

Moles water = 376.4 gms water over l8 gms/mole = 20.911 moles water

total moles = 21.680 moles

Mole fraction water = 20.911 moles water over 21.680 total moles = 0.9645

vapour pressure = 0.0313*0.9645 = 0.030189 atm

b). l8.3 gms solute over 88.8 gms water = X gms solute over l000 gms water (one kilogram)

set this up on a piece of paper and cross multiply, solving for X gms solute/kg water

X = 206.08 gms solute on one kilogram of water.

Now we can find the molality of this solution by this formula

molality = change temp boiling over boiling point constant for water

pure water boils at l00 degrees C but we are boiling l.4 degrees over that, so our change is l.4 degrees.

The boiling point constant for water is .52 degrees / molal.

molality = l.4 degrees/.52 degrees/molal = 2.69 molal

This tells us we have 2.69 moles of unknown in each kilogram of water.

And from above we know we also have 206.08 gms of solute per kilogram of water

Therefore, 2.69 moles of unknown must weigh 206.08 grams

so one mole unknown must weigh 206.08 gms over 2.69 moles = 76.5 grams

c).

Moles Cu(NO3)2 = 18.1 gms over 187.5 grams per mole = .09653 moles.

Each mole of copper nitrate will contain one mole Cu++, 2 moles of NO3- ions

so moles Cu++ inside .09653 moles of copper nitrate = 1 x .09653 = .09653 moles Cu++

moles NO3 = 2 x .09653 = 0.193 moles

Total moles ions = 0.193 + 0.09653 = 0.2895 moles