What mass of ammonium sulfate ((NH4)2SO4 132.15 g/mol) is required to produce 282 mL of a 2.50M solution of ammonium sulfate?

93.17 g

Explanation:

Recall that: mol = mass/molar mass

Also;               mol = molarity x volume

mole of (NH4)2SO4 to be prepared = 282/1000 (dm3) x 2.50 (mol/dm3)

                                                             = 0.705 mol

This can be used to determine the mass of (NH4)2SO4 that will be required.

mass = mole x molar mass

Hence, mass of (NH4)2SO4 required = 0.705 x 132.15

                                                               = 93.17 g

Hence, the mass of ammonium sulfate that will be required to prepare 282 mL of a 2.50M solution is 93.17 g

i think the answer is terrae but not 100 percent sure. ▓▬▬ §