A solution with a total volume of 850.0 mL contains 43.5 g Mg(NO3)2. If you remove 25.0 mL of this solution and then dilute this 25.0 mL sample with water until the new volume equals 600.0 mL, what is the concentration of nitrate ions in the 600.0 mL solution?

A solution with a total volume of 850.0 mL contains 43.5 g Mg(NO3)2. If you remove 25.0 mL of this solution and then dilute this 25.0 mL sample with water until the new volume equals 600.0 mL, what is the concentration of nitrate ions in the 600.0 mL solution?
a. 0.345M
b. 0.880M
c. 0.0288 M
d. 0.0122M
e. 0.0144M

Answer

• 0.0144M
  Explanation:
  Concentration of 850.0 mL Mg(NO3)2 solution
  moles / volume
  moles = 43.5 g / 148.3 g/mol
  = 0.293 moles
  Molarity = 0.293 moles / 0.85 L = 0.345 M
  Therefore, Molarity of 25.0 mL solution will also be 0.345 M
  Formula = M1V1 = M2V2
  M1 = Molarity of 25 mL solution = 0.345 M
  V1 = Volume = 25 mL
  M2 = Molarity of 600 mL solution = ?
  V2 = Volume = 600 mL
  0.345 * 25 = M2 * 600
  M2 =( 0.345 * 25) / 600
  M2 = 0.0144 M
  The molarity of 600 mL solution is 0.0144M