Description
Suppose 2.25 g of sodium chloride is dissolved in 50mL of a .20M aqueous solution of ammonium sulfate.
Calculate the final molarity of sodium cation in the solution. You can assume the volume of the solution doesn't change when the sodium chloride is dissolved in it.Round your answer to 2 significant digits.Explanation & Answer
initial mole of NaCl = 0.20M x 50x10^-3 L = 0.01 mol
molar mass of NaCl = 58.44 g/mol
additional mole = 2.25g / 58.44 g/mol = 0.0385 mol
final mole = 0.01 mol + 0.0385 mol = 0.0485 mol
since mole of NaCl = mole of Na+
The final molarity of sodium cation in the solution is then calculated by:
0.0485 mol / 50x10^-3 L = 0.97 M
Answer: 0.97 M