Answer to Question #206126 in Chemistry for Taylor

Question

1. What is the concentration, in g/mL, of a solution made by dissolving 75.0 g of phosphoric acid in 50.0 mL of solution?
2. What mass of NaCl is needed to produce 500 mL of a 2.0 mol/L solution?
3. What is the molarity of a solution that contains 0.350 g of barium fluoride, BaF₂, in 0.50 L of solution?
4. What is the molar concentration of a solution containing 10.0 g of NaOH in 250 mL of solution?
5. What mass of silver nitrate, AgNO₃, is needed to make 0.50 L of a 0.10 mol/L solution?
6. What volume of 1.0 mol/L solution of CuS can be made from 23.9 g of CuS?
7. A 100 mL sample of a 6.0 mol/L solution of H₂SO₄ is diluted to a final volume of 500 mL. What is the concentration of the diluted solution?
8. What volume of a 6.0 M NaOH solution is needed to produce 3.0 L of a 0.1 M solution?
9. Explain the steps you would follow to make 500 mL of a 0.5 mol/L solution of sodium chloride (NaCl) in the lab.

Answer 1

Solution:

(1):

phosphoric acid - H₃PO₄

(75.0 g H₃PO₄) × (1 / 50.0 mL) = 1.5 g/mL H₃PO₄

(2):

The molar mass of NaCl is 58.5 g/mol.

Thus,

(2 mol NaCl / 1 L) × (500 mL) × (1 L / 1000 mL) × (58.5 g NaCl / 1 mol NaCl) = 58.5 g NaCl

(3):

The molar mass of BaF₂ is 175 g/mol.

Thus,

(0.350 g BaF₂) × (1 mol BaF₂ / 175 g BaF₂) × (1 / 0.50 L) = 0.004 mol/L BaF₂

(4):

The molar mass of NaOH is 40 g/mol.

Thus,

(10.0 g NaOH) × (1 mol NaOH / 40 g NaOH) × (1 / 250 mL) × (1000 mL / 1 L) = 1.0 mol/L NaOH

(5):

The molar mass of AgNO₃ is 170 g/mol.

Thus,

(0.10 mol AgNO₃ / 1 L) × (0.50 L) × (170 g AgNO₃ / 1 mol AgNO₃) = 8.5 g AgNO₃

(6):

The molar mass of CuS is 95.6 g/mol.

Thus,

(23.9 g CuS) × (1 mol CuS / 95.6 g CuS) × (1 L / 1.0 mol CuS) = 0.25 L of solution

(7):

The equation for dilution calculations can be used.

where:

c₁, V₁ - the molarity and volume of the concentrated solution

c₂, V₂ - the molarity and volume of the diluted solution

Thus:

(6.0 mol/L) × (100 mL) = c₂ × (500 mL)

c₂ = (6.0 mol/L × 100 mL) / (500 mL) = 1.2 mol/L

c₂ = 1.2 mol/L

(8):

The equation for dilution calculations can be used.

where:

c₁, V₁ - the molarity and volume of the concentrated solution

c₂, V₂ - the molarity and volume of the diluted solution

Thus:

(6.0 M) × V₁ = (0.1 M) × (3.0 L)

V₁ = (0.1 M × 3.0 L) / (6.0 M) = 0.05 L

V₁ = 0.05 L

(9):

The molar mass of NaCl is 58.44 g/mol.

Thus,

(0.5 mol NaCl / 1 L) × (58.44 g NaCl / 1 mol NaCl) × (500 mL) × (1 L / 1000 mL) = 14.61 g NaCl

Steps of preparation of a 0.5 mol/L solution of sodium chloride (NaCl):

1. 14.61 grams of sodium chloride (NaCl) is weighed out and added to a 500 mL volumetric flask that has been about half-filled with distilled water.

2. The solution is swirled until all of NaCl dissolves.

3. More distilled water is carefully added up to the line etched on the neck of the flask (up to 500 mL).

4. The flask is capped and inverted several times to completely mix.

Answers:

1) 1.5 g/mL H₃PO₄

2) 58.5 g NaCl

3) 0.004 mol/L BaF₂

4) 1.0 mol/L NaOH

5) 8.5 g AgNO₃

6) 0.25 L of solution

7) c₂ = 1.2 mol/L

8) V₁ = 0.05 L

9) 14.61 g NaCl are dissolved in 500 mL volumetric flask.