Question

In: Chemistry

Which of the following four aqueous solutions will have the lowest freezing point? i) 0.05 m...

Which of the following four aqueous solutions will have the lowest freezing point? i) 0.05 m sucrose ii) 0.05 m CaCl2 iii) 0.05 m NaCl iv) 0.05 m AlCl3

MgCl2 is widely used to prevent ice formation on roadways. If this solution has non-ideal behavior, the van’t Hoff factor for a 0.05 m aqueous solution of MgCl2 would reasonably be _____. a. 1 b. 2 c. 2.7 d. 3 e. 3.2

Assuming ideal behavior, the change in the freezing point (deltaTf) for a 0.171m solution of NaCl in an unknown polar solvent is found to be −6.81 °C. Again, assuming ideal behavior, the change in the freezing point for a 0.684 m solution of NaCl in the same polar solvent will be ________.

a. −27.24 °C b. −6.81 °C c. +6.81 °C d. −54.48 °C e. −13.62 °C

Solutions

Expert Solution

Given aqueous solutions are having the same concentration that is 0.05m.

i) 0.05 m sucrose:

Sucrose is a covalent molecule, hence it won't undergo dissociation.

So there is no change in concentration.

ii) 0.05 m CaCl2

CaCl2(aq) -> Ca+2 (aq)+ 2Cl-(aq)

Hence, three ions per molecule are produced.

iii) 0.05 m NaCl

NaCl(aq) -> Na+(aq) + Cl-(aq)

Hence, two ions per molecule are produced.

iv) 0.05 m AlCl3

AlCl3(aq) -> Al+3(aq) +3Cl-(aq)

Hence, four ions per molecule are produced.

Lowering in freezing point depends on the number of solute particles.

option iv) 0.05m AlCl3 will have lowest freezing point.

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MgCl2 undergoes dissociation in aqueous solution.

MgCl2(aq) -> Mg+2 (aq)+ 2Cl-(aq)

vont hoff factor i = 1+ alpha(n-1) ----------------(1)

where alpha =degree of dissociation (in case of ionic compound, alpha=100%)

n=number of ions produced on dissociation

Substitute in (1), we get the value of i.

i=1+ (3-1)

therefore i=3

option d is the correct answer.

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deltaTf = Kf x molality

Given deltaTf1 =−6.81 °C

molality1 = 0.171m

When molality2 = 0.684 m

What is the value of deltaTf2 ?

Kf is constant for a particular solvent. So deltaTf1 /molality1 = deltaTf2/molality2

=> −6.81 °C/ 0.171m = deltaTf2/0.684 m

=> deltaTf2 = (−6.81 °C/ 0.171m ) . 0.684 m

                   = -27.24°C

Hence, option a is the correct answer.


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