The following experiment was conducted i) An evaporating dish was weighed and th

Question

The following experiment was conducted i) An evaporating dish was weighed and the mass is 257.4167 g ii) 100. 0 mL of a unknown acetate MW 142.39 g/mol, solution is placed in the dish and heated to dryness A white crystalline solid is observed in the dish iii) the evaporating dish is weighed again and the mass is 257.5591 g Determine the molarity of the unknown magnesium acetate solution Explain how you would use solid magnesium acetate to prepare 50.0mL of a stock solution that is ten times as concentrated as the unknown solution? Explain how you would dilute the stock solution prepared in part b to make 100 mL of the unknown solution to use for the experiment in part a.

Explanation / Answer

a. molality of Mg acetate solution

= (257.5591 - 257.4167)g/142.39 g/mol x 0.1 L

= 0.01 M

b. To prepare a 50 ml of 0.1 M Mg acetate solution,

grams of Mg acetate needed = 0.1 M x 0.05 L x 142.39 g/mol

                                               = 0.712 g to be dissolved in 50 ml water

c. To make 100 ml of 0.01 M solution from 0.1 M solution

volume of 0.1 M solution needed = 0.01 m x 100 ml/0.1 M = 10 ml

So mix 10 ml of 0.1 M solution with 90 ml water to get 100 ml of 0.01 M solution

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