A)When 4.52 g of a nonelectrolyte solute is dissolved in water to make 795 mL of

Question

A)When 4.52 g of a nonelectrolyte solute is dissolved in water to make 795 mL of solution at 25 °C, the solution exerts an osmotic pressure of 871 torr. Find the concentration, moles of solute and mass of solute.

B)A solution is made by dissolving 0.683 mol of nonelectrolyte solute in 809 g of benzene. Calculate the freezing point, Tf, and boiling point, Tb, of the solution.

C)Assuming 100% dissociation, calculate the freezing point and boiling point of 1.12 m AgNO3(aq).

D)For each solute, identify the better solvent: water or carbon tetrachloride: Br2, CaCl2, C6H6, CH3OH

Explanation / Answer

A)When 4.52 g of a nonelectrolyte solute is dissolved in water to make 795 mL of solution at 25 °C, the solution exerts an osmotic pressure of 871 torr. Find the concentration, moles of solute and mass of solute.

Solution :-

T = 25 C +273 = 298 K

Pressure = 871 torr * 1 atm / 760 torr =1.146 atm

Using the osmotic pressure lets calculate the concentration of the solution.

Pi = MRT

Where Pi = osmotic pressure in atm

M= molarity

R= constant 0.08206 L atm per mol K

T= Kelvin temperature

Now lets put the values in the formula and find molarity

1.146 atm = M * 0.08206 L atm per mol K * 298 K

1.146 atm / 0.08206 L atm per mol K * 298 K = M

0.04686 mol per L = M

Now lets calculate the moles of the solute using the molarity and volume

Moles = molarity * volume in liter

Moles = 0.04686 mol per L * 0.795 L

            = 0.037254 mol

Now lets calculate the molar mass of the solute

Molar mass = mass / moles

                    = 4.52 g / 0.037254 mol

                    = 121.33 g per mol

B)A solution is made by dissolving 0.683 mol of nonelectrolyte solute in 809 g of benzene. Calculate the freezing point, Tf, and boiling point, Tb, of the solution.

Solution :-

Lets first calculate the molality of the solution using the moles and kg of the solvent

809 g * 1 kg / 1000 g = 0.809 kg

Molality = moles / kg solvent

Molality = 0.683 mol / 0.809 kg

                = 0.8443 m

Now lets calculate the freezing point of the solution

Molal freezing point of the benzene = 5.12 C/m

Freezing point of the pure benzene = 5.5 C

Delta Tf = Kf* m

              = 5.12 C/m * 0.8443 m

              = 4.32 C

Therefore freezing point of solution = freezing point of pure solvent – delta Tf

                                                           = 5.5 C – 4.32 C

                                                          = 1.18 C

Now lets calculate the boiling point of the solution

Molal boiling point constant of the benzene Kb = 2.65 C/m

Boiling point of benzene = 80.1 C

Delta Tb = Kb * m

                 = 2.65 C/m * 0.8443 m

                 = 2.24 C

Therefore boiling point of the solution = B.P of pure benzene + delta Tb

                                                               = 80.1 C + 2.24 C

                                                               = 82.34 C

C)Assuming 100% dissociation, calculate the freezing point and boiling point of 1.12 m AgNO3(aq).

Solution :-

On the dissociation the AgNO3 will form two ions as Ag^+ and NO3^-

So the total ion concentration = 1.12 m * 2 = 2.24 m

Now lets calculate the freezing point of solution.

Kf of water = 1.86 C/m and kb of water = 0.512 C/m

Delta Tf = Kf*m

               = 1.86 C/m * 2.24 m

               = 4.17 C

Freezing point of solution = freezing point of water – delta Tf

                                             = 0 C – 4.17 C

                                              = -4.17 C

Now lets calculate the boiling point of the solution

Delta Tb = Kb*m

               = 0.512 C/m * 2.24 m

              = 1.15 C

Boiling point of solution = boiling point of water + delta Tb

                                         = 100 C + 1.15 C

                                         = 101.15 C

D)For each solute, identify the better solvent: water or carbon tetrachloride: Br2, CaCl2, C6H6, CH3OH

Solution :-

Polar solute dissolves in polar solvents and non polar solute dissolves in non polar solvents

Water is polar solvent and carbon tetrachloride is non polar solvent

Therefore

CaCl2 and CH3OH will dissolve in water because they are polar

C6H6 and Br2 will dissolve in carbon tetrachloride because they are non polar.

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