Question

Suppose 1.89g of ammonium chloride is dissolved in of a 76.0mm aqueous solution of potassium carbonate. Calculate the final molarity of ammonium cation in the solution?

Answer 1

To calculate the final molarity of ammonium cation in the solution, divide the moles of ammonium chloride by the volume of the solution.

Step-by-step explanation:

To calculate the final molarity of ammonium cation in the solution, we need to first calculate the moles of ammonium chloride and then divide by the volume of the solution.

The given mass of ammonium chloride is 1.89 g. To convert this to moles, we use the molar mass of ammonium chloride (NH4Cl), which is 53.49 g/mol. Therefore, the moles of NH4Cl are 1.89 g / 53.49 g/mol = 0.0353 mol.

The volume of the solution is given as 76.0 mL. To convert this to liters, we divide by 1000: 76.0 mL / 1000 mL/L = 0.076 L.

Finally, we can calculate the molarity of the ammonium cation (NH4+) by dividing the moles of NH4Cl by the volume of the solution in liters: 0.0353 mol / 0.076 L = 0.464 M.