Answer:
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
