Suppose of sodium chloride is dissolved in of a 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 significant digits.

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.

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