Question

What is the vapor pressure of a solution prepared by adding 50.0 g of NaCl to 250. g of pure water at 25 ^oC? (Water has a vapor pressure of 0.036atm at 25 ^oC)

0.006 atm

0.004 atm

0.030 atm

0.032 atm

0.034 atm

0.002 atm

Answer 1

Solution :-

Lets first calculate the moles of the NaCl and H2O

Moles = mass / molar mass

Moles of NaCl =52 g /58.443 g per mol =0.8555 mol NaCl

But NaCl forms 2 ions therefore total moles of ions = 0.8555 mol * 2 = 1.711 mol

Now lets calculate moles of water

Moles of water = 250 g / 18.0148 g per mol =13.877 mol

Now lets calculate the mole fraction of the water

Mole fraction = moles of water / total moles

                        = 13.877 mol / (13.877 mol + 1.711 mol)

                       = 0.8902

Now lets calculate the vapor pressure of the solution

Vapor pressure of solution = mole fraction of solvent * vapor pressure of pure solvent

                                                 = 0.8902 * 0.036 atm

                                                 = 0.032 atm