In: Chemistry
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
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.