Question

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1. Calculate the molarity, molality and mass % of NH3 in a solution in which 60.0g of NH3 are dissolved in 140.0g of water. The Density of this solution is 0.982g/mL

2. A solution is prepared by dissolving 38.0g of non-volatile molecular solute in 125 grams of benzene. The boiling point of this solution is 84.5C (Celsius) and the boiling point of pure benzene is 80.1C (Celsius). What is the molar mass of the solute. NOTE: The molal boiling point elevation constant (Kb) for benzene is 2.53C (Celsius) kg/mol

3. Calculate the osmotic pressure at 298K of an aqueous solution of CaF2 if 0.025 moles of this salt are dissolved and diluted to a volume of 500. mL of solution. (use R=0.0821L atm / mol K)

4. If liquid A has a vapor pressure of 176mmHg at 25 degrees Celsius

            a) What would be the vapor pressure of a solution prepared by combining 3.50 moles of B (Non-Volatile compound) and 17.50 of A

            b) If now we mix 6.60 moles of A and 3.50 moles of C (a liquid that has a vapor pressure of 135 mmHg at 25 degrees Celsius) what would be the vapor pressure of the mixture?

Answer 1

1. Calculate the molarity, molality and mass % of NH3 in a solution in which 60.0g of NH3 are dissolved in 140.0g of water. The Density of this solution is 0.982g/mL

Solution :- mass of solution = 140.0 g water +60.0 g NH3 = 200 g

Volume of solution = 200 g / 0.982 g per ml = 203.7 ml = 0.2037 L

Lets calculate the moles of the NH3

Moles of NH3 = 60.0 g / 17.03 g per mol

                         = 3.523mol NH3

Molarity = moles / liter of solution

                = 3.523 mol / 0.2037 L

                = 17.3 M

Molality = moles / kg solvent

             = 3.523 mol / 0.140 kg

           = 25.16 m

% by mass = (mass of solute / mass of solution)*100%

                    = ( 60.0 g / 200 g) *100 %

                   = 30.0 %

2. A solution is prepared by dissolving 38.0g of non-volatile molecular solute in 125 grams of benzene. The boiling point of this solution is 84.5C (Celsius) and the boiling point of pure benzene is 80.1C (Celsius). What is the molar mass of the solute. NOTE: The molal boiling point elevation constant (Kb) for benzene is 2.53C (Celsius) kg/mol

Solution :-

Change in the boiling point Delta Tb = 84.5C – 80.1 C = 4.4 C

Now lets calculate the molality

Delta Tb= Kb* m

4.4 C = 2.53 C per m * m

4.4 C / 2.53 C per m = m

1.739 m = m

So the molality of the solution is 1.739 m

Now lets calculate the moles of solute

Moles of solute = molality * kg solvent

                           = 1.739 mol per kg * 0.125 kg

                           = 0.21738 mol

Molar mass = mass in gram / moles

                   = 38.0 g / 0.21738 mol

                   = 174.8 g per mol

We can round it to 175 g /mol

3. Calculate the osmotic pressure at 298K of an aqueous solution of CaF2 if 0.025 moles of this salt are dissolved and diluted to a volume of 500. mL of solution. (use R=0.0821L atm / mol K)

Solution :-

Molarity = moles / liter

                = 0.025 mol / 0.500 L

               = 0.50 M

Osmotic pressure = i * M* R*T

                              = 3 * 0.50 mol per L * 0.0821 L atm per mol K * 298 K

                              = 36.7 atm

4. If liquid A has a vapor pressure of 176mmHg at 25 degrees Celsius

            a) What would be the vapor pressure of a solution prepared by combining 3.50 moles of B (Non-Volatile compound) and 17.50 of A

solution :- lets first calculate the mole fraction of A

mole fraction of A= moles of A / (moles of A + moles of B)

                              = 17.50 / (17.50 + 3.50 )

                             = 0.8333

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

                                           = 0.8333 * 176 mmHg

                                           = 146.7 mmHg

            b) If now we mix 6.60 moles of A and 3.50 moles of C (a liquid that has a vapor pressure of 135 mmHg at 25 degrees Celsius) what would be the vapor pressure of the mixture?

Solution :-

mole fraction of A= moles of A / (moles of A + moles of C )

                             = 6.60 / (6.60 + 3.50 )

                             = 0.6535

Mole fraction of C = 1-0.6535 = 0.3465

Vapor pressure of A = mole fraction A * vapor pressure of pure A

                                  = 0.6535 * 176 mmHg

                                  = 115 mmHg

Vapor pressure of C = mole fraction B * vapor pressure of pure B

                               = 0.3465 * 135 mmHg

                                 = 46.8 mmHg

Vapor pressure of solution = Vapor pressure of A + vapor pressure of C

                                             = 115 mmHg + 46.8 mmHg

                                            = 161.8 mmHg