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I would appreciate if someone can help me with detailed answer to this question please? Thank you very much!!

8. Magnesium acetate Mg (CH3coo option for road salt. )2 has been suggested as a less damaging g/ml A 3.85 M 1.47 magnesium acetate solution has a density of a) Given the data on the data page, determine its normal freezing point. b) If you had a solution sodium chloride with exactly the same molar of would you expect its normal freezing point to be higher or lower than the magnesium acetate solution? Explain briefly a few words is plenty) The reaction of aluminum oxide reacts with sodium metal in the reaction Al203 (s) Na(s) Na20 (s) A (s)

Explanation / Answer

a.

First calculate the molality m of a 3.85 M Mg(CH3COO)2 solution.

1 000 mL x 1.47 g/mL = 1470 g, solution mass

3.85 mol/L x1 L = 3.85 mol

3.85 mol x 142.4 g/mol = 548 g Mg(CH3COO)2 , solute mass

1470 g – 548 g = 1192 g solvent (water)

The molality is

3.85 mol/ 1.192 kg solvent = 3.23 mol/1000 k solvent = 3.23 m

The freezing point depression is

T = i Kf m

where i = number of particles per formula unit at dissolution (i = 3 for Mg(CH3COO)2 ), Kf is a specific constant for water (1.86 ºC/m), m is the solute molality.

T = 3x 1.86 ºC/m x 3.23 m = 18.0 ºC

Thus the freezing point of this solution is 0 ºC - 18 ºC = -18 ºC

b.

The only difference is i =2 for NaCl

For NaCl the freezing point will be -12 ºC   (higher).

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