Answer:
0.103 M.
Explanation:
The balanced equation of reaction is given below;
2NaCl + (NH4)2SO4 -----------------> Na2SO4 + 2NH4Cl
The parameters given from the question are; mass of Sodium chloride, NaCl = 0.197 g, the volume of ammonium sulphate is 100 mL = 0.1 litres, concentration of ammonium sulfate = 54.0mM = 0.054 mol/L.
Step one: Calculate the number of moles of NaCl by using the formula below;
Number of moles,n= mass/ mass number.
Number of moles,n = 0.917/ 58.5.
Number of moles, n= 0.0157 g/mol.
Step two: find the number of moles of ammonium sulfate.
Hence, the number of moles of ammonium sulfate, n = concentration (mol/L) of ammonium sulfate × volume.
Therefore, the number of moles of ammonium sulfate, n = 0.054 × 0.1.
the number of moles of ammonium sulfate, n = 0.0054 mol.
Step three: Calculate the excess moles of NaCl.
That is; 0.0157 - 0.0054 = 0.0103 mol.
Step four: Calculate the final molarity of sodium cation in the solution.
[Na^+] = number of moles/ total volume.
[Na^+] = 0.0103 mol/ 0.1.
0.0103 mol= 0.103 M.
