Question

2. Consider a 0.15 m strontium chloride aqueous solution.

a. Determine the molality (in m) of all solutes in the solution. Assume that strontium chloride completely ionizes when dissolved in water. All solutes means all cations and anions. The molality that you determine here is the value for msolute of (Equation 3) in the “Background and Procedure” file. You must explain your answer. No credit will be given if you write only calculations. (Hint: See Example 3 of the Background and Procedure file.) (10 points)

(Equation 3) ∆T= T_{pure solvent} - T_solution = K_f * m_solute (Equation 3)

It is important to remember that the molality of solutes used in Equation 3 refers to all solutes present in the solution. Example 3: When an ionic compound (strong electrolyte) such as CaCl2 dissolves in water, the compound would completely ionize into cations and anions because CaCl2 is a strong electrolyte. CaCl2(aq) → Ca2+(aq) + 2 Cl−(aq) For CaCl2 solution, every one mole of CaCl2 would produce three moles of solutes combining Ca2+ cations and Cl− anions. So if the concentration of CaCl2 solutions is 1.0 m, then the molality of all solutes in Equation 3 should be calculated by multiplying the concentration of CaCl2 by “3” as follows. 3 × 1.0 m = 3.0 m Again, the multiplier “3” in front of 1.0 m originates from the fact that one CaCl2 completely ionizes into “three” ions (one Ca2+ ion and two Cl− ions). The multiplier depends on ionic compounds. So do not assume that it is always “3”.

b. Consult Table 1. Select the molal freezing point depression constant for the solution of this question. Make sure to write the units! (Hint: What is the solvent in this question?)

Table 1: Molal Freezing Point Depression Constants (Kf) for Some Solvents1 Solvent Kf (°C⋅kg/mol) : Water 1.86 Toluene 3.55 Glycerol 3.56 Benzene 5.07

Kf = ____________

c. Determine the magnitude of the freezing point depression (∆T) of the solution. Show your calculation. (10 points) (Hint: Remember that m = mol/kg when m represents the molality.)

d. Determine the freezing point of the solution. The freezing point of water is 0°C exactly. Show your calculation. (10 points)

Answer 1

2. a. Given 0.15 m Strontium chloride solution (SrCl₂) which ionises completely as shown in solution

So 1 molecule of SrCl₂ gives 3 ions in solution.

So total molality of all solutes in given solution=3 x 0.15 m=0.45 m

b. As this is an aqueous solution, so the solvent is water.

Molal freezing point depression constant of water=K_f=1.86 (°C.kg/mol)

c. The freezing point depression ∆T=K_f​​​​​​xm

=1.86 °C/mx 0.45 m

=0.837 °C

d. Freezing point of water=0°C=T_{pure solvent}

Freezing point of solution=T_solution

Freezing point depression=∆T=T_{pure solvent} - T_solution

0.837°C=0°C-T _solution

T_solution=0°C-0.837°C=-0.837°C

So freezing point of the solution=-0.837°C