How many grams of ammonium sulfate are needed to make a 0.25 L solution at a concentration of 6 M?

Moles = mass of solute/ gram molecular mass.
Solute is the 'stuff being disolved.
Solution is a homogeneous mixture of a solute in a solven. G.M.M. is
gram molecular mass

. moles of solute

Molarity
is defined as: ___________________________________

. liters of
solution

.
[g
(NH4)2SO4/G.M.M]

6
M = ______________________________

. L
(NH4)2SO4(aq)

.
? g
/
132g/mol

6 M = __________________

.
0.25 L

g
(NH4)2SO4
= 6 mol/L * 0.25 L * 132 g/mol = 198
g(NH4)2SO4

Since
6M is only 1 significant figure the answer to 1 significant figure
would be 200 g
(NH4)2SO4.
6.0M solution would also use 200 (with the bold numbers representing
the significant figures)

If the solution were 6.00M (3 sig figs)
then you would need 198
g(NH4)2SO4