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 =
since mole of NaCl = mole of Na+
The final molarity of sodium cation in the solution is then calculated by:
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