Question

Suppose 18.2 g of ammonium acetate is dissolved in 250. mL of a 0.50 M aqueous solution of sodium chromate. Calculate the final molarity of ammonium cation in the solution. You can assume the volume of the solution doesn't change when the ammonium acetate is dissolved in it. Round your answer to 3 significant digits.

Answer 1

CH3COONH4 < === > CH3COO- + NH4+

Number of moles = amount in g / molar mass

= 18.2 g/ 77.08248 g/mol

= 0.234 moles of CH3COONH4 or NH4+

Volume = 250 ml or 0.250 L

Molarity = number of moles / volume in L

= 0.234 Moles / 0.250 L

= 0.936 M