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Question

What is the procedure and solution in performing these?

1. How many mL of 0.310 M barium nitrate are required to precipitate as barium sulfate all the sulfate ions from 28.0 mL of 0.380 M aluminum sulfate? (Stoichiometry)
2. How many grams of silver chromate will precipitate when 190 mL of 0.900 M silver nitrate are added to 140 mL of 0.800 M potassium chromate? (Stoichiometry)
3. What is the freezing point of a solution of 15.0 g of NaCl in 250 g of water? (Colligative)

Answer 1

1). Write the balanced equation: 3Ba(NO₃)₂ + Al₂(SO₄)₃ ==> 3BaSO₄(s) + 2Al(NO₃)₃

2). Find mols of Al₂(SO₄)₃: 28.0 ml x 1 L/1000 ml x 0380 mol/L = 0.0106 mols Al₂(SO₄)₃

3). Find mols of Ba(NO₃)₂ needed: 0.0106 mols Al₂(SO₄)₃ x 3 mols Ba(NO₃)₂ /mol Al₂(SO₄)₃ = 0.0319 mols

4). Find volume Ba(NO₃)₂: 0.0319 mols Ba(NO₃)₂ x 1 L/0.310 mols = 0.103 liters = 103 mls Ba(NO₃)₂

1). Balanced equation: K₂CrO₄ + AgNO₃ ==> Ag₂CrO₄(s) + 2KNO₃

2). mols AgNO₃ = 190 ml x 1 L/1000 ml x 0.900 mol/L = 0.171 mol AgNO₃

3). mols K₂CrO₄ = 140 ml x 1L/1000 ml x 0.800 mol/L = 0.112 mol K₂CrO₄ = LIMITING REACTANT

4). grms Ag₂CrO₄ = 0.112 mol K₂CrO₄ x 1 mol Ag₂CrO₄/mol K₂CrO₄ x 332 g Ag₂CrO₄/mol = 37.2 g Ag₂CrO₄

1). Use ∆T = imK where ∆T is change in freezing point; i = 2 for NaCl; m = molality (see below); K=freezing point constant for water = 186º/m

2). Find m: 15.0 g NaCl x 1 mol NaCl/58.4 g = 0.257 mols / 0.250 kg = 1.03 molal

3). Solve for ∆T: ∆T = (2)(1.03)(1.86) = 3.83º

4), New freezing point = 0º - 3.83º = -3.83ºC