Question

How many moles of (NH4)2Fe(SO4)2 are in 94.00 mL of 2.50 M (NH4)2Fe(SO4)2 solution.

Answer 1

`n_(solute)=0.235mol`

Step-by-step explanation:

Hello.

In this case, we are talking about molarity which is an unit of concentration relating the moles of the solute and the volume of the solution in liters only:

`M=(n_(solute))/(V_(solution))`

Since the solute is the (NH4)2Fe(SO4)2 and the volume in liters:

`V=94.00mL*(1L)/(1000mL) =0.094L`

Thus, for the given 2.50-M solution, the moles of solute result:

`n_(solute)=M*V=2.50mol/L*0.094L\\\\n_(solute)=0.235mol`

Best regards.