The following experiment was conducted: i. An evaporating dish was weighed and t

Question

The following experiment was conducted: i. An evaporating dish was weighed and the mass is 2574167 g ii. 100.0 ml, of a unknown magnesium acetate, MW 142.39 g/mol, solution is placed in the evaporating dish and heated to dryness. white crystal ne solid dish. iii. The evaporating dish is weighed again and the mass is 257.5591 Determine the molarity of the unknown magnesium acetate solution. b. Explain how you would use solid magnesium acetate to prepare 50.0 mL of a stock solution that is ten times as concentrated as the unknown solution? c. Explain how you dilute the stock solution prepared in part b to make would b 100 ml of the unknown solution to use for the experiment in part a.

Explanation / Answer

a)

M of MgAcetate

m = 257.4167 g of dish

total m = 257.5591 g

mass of acetate = 257.5591 -257.4167 = 0.1424 g

mol = masS/MW = 0.1424 / 142.39 = 0.0010000 mol

M = mol/V = (0.0010000)/(100*10^-3) = 0.010 M of MGAcetate

b)

if we required V = 50 mL of stock solution, which is 10 times more concentrated

M = 10*0.010 = 0.1 M

now, we need V = 50 mL

mol = MV= 50*0.1 = 5 mmol

mass = mmol*MW = 5*142.39 = 711.95 mg = 0.71195 g

add 0.71195 g must be added to 50 mL

c=

dilute to 100 mL

M1*V1 = M2*V2

M1*V1 = 1/10*M1 * V2

V1 = 1/10*100

V1 = 10 mL

take 10 mL and dilute it up to V = 100 mL

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