Question

Given the following aqueous solutions answer the following questions, briefly explain your answer and show all calculations to support your answer: 0.050 m sodium chloride 0.050 m C6H12O6 (glucose in water) 0.0050 m carbon dioxide 0.050 m ammonium carbonate

a. How much of each solute would you need to make 0.500 L aqueous solutions of the compounds listed above? b. Give that at 25oC water has an approximate density of 1.00 g/mL, what is the molarity of each of the solutions listed above? c. Which solution has the largest van’t Hoff factor? Which has the smallest? d. Which solution above depends heavily on the pressure of the system to keep the concentration constant? e. Which solution has the highest freezing point? f. Which solution has the highest boiling point?

Answer 1

The aqueous solutions given in the question are:

0.050 m sodium chloride

0.050 m glucose

0.0050 m carbon dioxide

0.050 m ammonium carbonate

a. How much of each solute would be needed to make 0.500 L aqueous solutions of the compounds listed above

As, molality = no. of moles of solute / mass of solvent in kg

Making 0.500 L aqueous solution of 0.050 m sodium chloride

As number of moles, n = mass of solute / molar mass

Therefore, molality = n of solute / mass of solvent in kg

m = (solute mass in g / solute molar mass) / mass of solvent in kg

solute mass / solute molar mass = molality x mass of solvent in kg

solute mass in g = molality x mass of solvent in kg x solute molar mass ...............................(i)

Putting the values in the eq (i)

mass of sodium chloride = 0.05 x 0.5 x 58.4 = 1.46 g

So, mass of sodium chloride needed to make 0.500 L aqueous solution = 1.46 g

Making 0.500 L aqueous solution of 0.050 m glucose

Putting the values in the eq (i)

mass of glucose = 0.05 x 0.5 x 180 = 4.5 g

Therefore, mass of glucose needed to make 0.500 L aqueous solution = 4.5 g

Making 0.500 L aqueous solution of 0.050 m carbon dioxide  

Putting the values in the eq (i)

mass of glucose = 0.05 x 0.5 x 44 = 1.1 g

So, mass of carbon dioxide needed to make 0.500 L aqueous solution = 1.1 g

Making 0.500 L aqueous solution of 0.050 m ammonium carbonate  

Putting the values in the eq (i)

mass of glucose = 0.05 x 0.5 x 96.1 = 4.5 g = 2.4 g

So, mass of ammonium carbonate needed to make 0.500 L aqueous solution = 2.4 g

(b) Molarity of each of the solutions listed above

Molarity (M) is defined as the number of moles of solute dissolved per litre of solution whereas, molality (m) is defined as the no. of moles of solute dissolved per kg of solvent.

As it is given that  at 25^oC water has an approximate density of 1.00 g/mL. This means that the volume of the solvent (water) in L is equal to the mass of it in kg. So, 0.5 L of water is equal to 0.5 kg of water. Therefore, the values of molarity of each of the solutions listed above would be approximately same as that of the molality.

(c) Which solution has the largest van’t Hoff factor? Which has the smallest?

The van't Hoff factor, i, can be defined as the number of particles formed in a solution from one formula unit of solute.

it is a property of the solute and in an ideal solution, it does not depend on the concentration of the solution.

van’t Hoff factor of sodium chloride

NaCl(s)→Na⁺(aq)+Cl^-(aq)

As one formula unit of sodium chloride gives two ions in the aqueous solution, so van’t Hoff factor of sodium chloride is 2.

van’t Hoff factor of glucose

As one formula unit of glucose does not dissociate into ions in the aqueous solution, so van’t Hoff factor of glucose is 1.

van’t Hoff factor of carbon dioxide

As one formula unit of glucose does not dissociate into ions in the aqueous solution, so van’t Hoff factor of carbon dioxide is 1.

But, aqueous solution of carbon dioxide may also contain traces of carbonic acid (H₂CO₃) which will give the van’t Hoff factor of 3.

van’t Hoff factor of ammonium carbonate

(NH₄)₂CO₃(s)→2NH₄⁺(aq)+CO_3^2-(aq)

As one formula unit of sodium chloride gives three ions in the aqueous solution, so van’t Hoff factor of ammonium carbonate is 3.

So, ammonium carbonate solution has the largest van’t Hoff factor value of 3 whereas glucose solution has the smallest of 1.

(d) Which solution above depends heavily on the pressure of the system to keep the concentration constant?

aqueous solution of 0.050 m carbon dioxide depends heavily on the pressure of the system to keep the concentration constant because gaseous carbon dioxide may get released out of the solution otherwise.

(e) Which solution has the highest freezing point?

It is known that van’t Hoff factor affects the freezing point of a solution. The solution having highest value of van’t Hoff factor will have lowest freezing point. So, aqueous ammonium carbonate solution which has the largest van’t Hoff factor value will have lowest freezing point whereas, aqueous glucose solution which has the lowest value of van’t Hoff factor, will have the highest freezing point.

(f) Which solution has the highest boiling point?

van’t Hoff factor also affects the boiling point of a solution. The solution having highest value of van’t Hoff factor will have highest boiling point. So, aqueous ammonium carbonate solution which has the largest van’t Hoff factor value will have highest boiling point.