Answer to Question #109431 in Organic Chemistry for Rosie

Question

1. What is the molarity of a solution that contains 10.0 grams of Silver Nitrate that has been dissolved in 750.0 mL of water?
2. You want to create a 0.25 M Potassium Chloride solution. You have mass of 5.00 grams of Potassium Chloride. How much water is needed?
3. What is the molality of a solution that contains 48 grams of Sodium Chloride and 250 mL of water?
4. What is the percentage mass of problem #1?
5. How many mL of hydrogen peroxide are needed to make an 8.50% solution by volume of hydrogen peroxide if you want to make 450.0 mL of solution?
6. What is the mole fraction of the solute in the solution of problem #1?
7. What is the mole fraction of the solvent in the solution of problem #1?
8. What is the molality ions in the solution of problem #3?
9. What is the molality of a solution that contains 13.4 grams of calcium chloride dissolved in 655.0 mL of water?
10. What is the molality ions in the solution of problem #9?

Answer 1

1. C = (m / M) / (V / 1000) = (10 / 170) / (750 / 1000) = 0.078 M

2. n(KCl) = 5 / 74.5 = 0.067 mol

V(H₂O) = 1000 × 0.067 / 0.25 = 268 mL

3. C(NaCl) = (48 / 58.5) / (250 / 1000) = 3.28 M

4. m_% = 100 × m(AgNO₃) / m_sol = 100 ×10 / (750 + 10) = 1.32 %

5. V(H₂O₂) = 450 × (8.5 / 100) = 38.25 mL

6. w(AgNO₃) = 100 × n(AgNO₃) / n_sol = 100 × (10 / 170) / (10 / 170 + 750 / 18) = 0.15 %

7. wH2O = 100 - wAgNO3 = 100 - 0.15 = 99.85 %

8. b = n(NaCl) / m(H₂O) = (48 / 58.5) / (250 / 1000) = 3.28 (mol / kg) (for both Na⁺ and Cl^-)

9. C(CaCl₂) = (13.4 / 111) / (655 / 1000) = 0.18 M

10. b(Ca²⁺) = 0.18 (mol / kg); C(Cl^-) = 2 × C(CaCl₂) = 0.36 (mol / kg)