Match the following aqueous solutions of each set with the appropriate letter fr
ID: 494196 • Letter: M
Question
Match the following aqueous solutions of each set with the appropriate letter from the column on the right.
1. 0.18 m CaBr2 A. Highest boiling point
2. 0.19 m CoI2 B. Second highest boiling point
3. 0.17 m Ca(NO3)2 C. Third highest boiling point
4. 0.59 m Sucrose (nonelectrolyte) D. Lowest boiling point
(use same A > B > C > D scale)
1. 0.14 m Fe(CH3COO)3
2. 0.25 m KNO3
3. 0.18 m CuCl2
4. 0.59 Sucrose
(use same scale)
1. 0.28 m NaCH3COO
2. 0.20 m CuBr2
3. 0.17 m Ba(OH)2
4. 0.53 m Ethylene glycol (nonelectrolyte)
Explanation / Answer
Ans. Boiling point elevation, dTb = i Kb m - equation 2
where, i = Van’t Hoff factor = 1 for non-dissociating solutes
Kb = molal boiling point elevation constant of the solvent
m = molality of the solution
dTf = boiling point elevation
#A. 1. dTb = 3 x Kb x 0.18 m = 0.54 Kb m ; [i = 3]
2. dTb = 3 x Kb x 0.19 m = 0.57 Kb m ; [i = 3]
3. dTb = 3 x Kb x 0.17 m = 0.51 Kb m ; [i = 3]
4. dTb = 1 x Kb x 0.59 m = 0.59 Kb m ; [i = 1]
Note: All dTb values have “Kb m” in common. So, compare the numerical values only. the greater is the numerical; value, greater is elevation in boiling point.
Order of boiling point (highest to lowest) = 4 > 2 > 1 > 3
#B. 1. dTb = 4 x Kb x 0.14 m = 0.56 Kb m ; [i = 4]
2. dTb = 2 x Kb x 0.25 m = 0.50 Kb m ; [i = 2]
3. dTb = 3 x Kb x 0.18 m = 0.54 Kb m ; [i = 3]
4. dTb = 1 x Kb x 0.59 m = 0.59 Kb m ; [i = 1]
Order of boiling point (highest to lowest) = 4 > 1 > 3 > 2
#C. 1. dTb = 2 x Kb x 0.28 m = 0.56 Kb m ; [i = 2]
2. dTb = 2 x Kb x 0.20 m = 0.60 Kb m ; [i = 3]
3. dTb = 3 x Kb x 0.17 m = 0.51 Kb m ; [i = 3]
4. dTb = 1 x Kb x 0.53 m = 0.53 Kb m ; [i = 1]
Order of boiling point (highest to lowest) = 2 > 1 > 4 > 3
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