Question

Please use sig figs and dimensional analysis and explain process!

1. Dissolve 69.0 g NaNO₂ in 1.00 x 10² mL H₂O.

(a) If the temperature of the solution is 100°C, what is the vapor pressure above the solution? The density of H₂O at 100°C is 0.9591 g/mL.

(b) Calculate the change in the boiling point of the solution, and calculate the boiling point of the aqueous solution.

(c) Calculate the freezing point of the solution. The density of H₂O at 0°C is 0.9998 g/mL.

(d) Which change is greater, the boiling point elevation or the freezing point depression?

(e) Estimate the osmotic pressure at 25 °C for the solution. Assume that dissolved sodium nitrite does not affect the volume of the solution.

Answer 1

1. Dissolve 69.0 g NaNO₂ in 1.00 x 10² mL H₂O.

(a) If the temperature of the solution is 100°C, what is the vapor pressure above the solution? The density of H₂O at 100°C is 0.9591 g/mL.

Solution :- to calculate the vapor pressure above the solution we need to calculate the ole fraction of the solvent

Because the formula to calculate the vapor pressure is as follows

Vapor pressure of solution = mole ffraction of solvent * vapor pressure of the pure solvent.

So lets first calculate the moles of the NaNO2 and moles of water

Moles = mass / molar mass

Moles of NaNO2 = 69.0 g / 68.9953 g per mol = 1.00 mol NaNO2

Moles of water = 95.91 g / 18.0148 g per mol =5.32 mol H2O

NaNO2 is the ionic salt so it gives 2 ions so total moles of ions = 1.00 * 2 = 2.00

Now lets calculate the mole fraction of water

Mole fraction of water = moles of water / (moles of water + moles of ions)

Mole fraction of water = 5.32 mol / (5.32 mol + 2.00mol)

                                      = 0.727

Now lets use this mole fraction of water to calculate the vapor pressure

Vapor pressure of pure water at 100 C = 760 mmHg

So, vapor pressure of the solution = 0.727 * 760 mmHg

                                                        = 553 mmHg

(b) Calculate the change in the boiling point of the solution, and calculate the boiling point of the aqueous solution.

Solution :- We already calculated the moles of the NaNO2 that is 1.00 and total moles of ions = 2.00

So lets calculate the molality of the solution

Molality = moles / kg solvent

Molality total ion = 2.00 mol / 0.100 kg                            (100 ml water = 100 g = 0.100 kg)

                             = 20.0 m

Now lets calculate the change in the boiling point of the solution

Formula

∆Tb = Kb* m

Where ∆Tb = change in boiling point , Kb=0.512 ^oC/m is the molal boiling point constant , m = molality

Lets put the values in the formula.

∆Tb = 0.512 ^oC per m * 20.0

∆Tb = 10.24 ^oC

So the change in the boiling point = 10.24 ^oC

Now lets calculate the boiling point of the solution

Boiling point of solution = Boiling point of pure solvent + ∆Tb

                                         = 100 ^oC + 10.24 ^oC

                                         =110.24 ^oC

(c) Calculate the freezing point of the solution. The density of H₂O at 0°C is 0.9998 g/mL.

Solution :-

Moles of total ions = 2.00

Mass of water = 99.98 g since density is 0.9998 g/ml

Molality of total ion = 2.00 mol / 0.09998 kg = 20.0 m

Formula to calculate the freezing point change is as follows

∆Tf = Kf* m

Where ∆Tf = change in freezing point , Kf=1.86^oC/m is the molal freezing point constant , m = molality

Lets calculate the change in freezing point

∆Tf = Kf* m

∆Tf = 1.86 C/m * 20.0 m

∆Tf = 37.2 ^oC

So change in the freezing point = 37.2 ^oC

Now lets calculate the freezing point of the solution

Freezing point of solution = freezing point of pure solvent - ∆Tf

                                           = 0.00 ^oC - 37.2 ^oC          (0.00 ^oC is the freezing point of pure water)

                                           = -37.2 ^oC

(d) Which change is greater, the boiling point elevation or the freezing point depression?

Solution :- Looking at the values of the ∆Tb and ∆Tf we can conclude that the change in the freezing point is greater.

(e) Estimate the osmotic pressure at 25 °C for the solution. Assume that dissolved sodium nitrite does not affect the volume of the solution.

Solution

Formula to calculate the osmotic pressure is as follows

∏ = MRT

Where ∏ = osmotic pressure , R= 0.08206 L atm per K mol , T= Kelvin temperature (25+273 = 298 K)

M= molarity

Lets put the values in the formual

∏ = 20.0 M * 0.08206 L atm per K mol * 298 K

∏ = 489 atm