How many grams of [tex]\left( NH_4 \right)_2 SO_4[/tex] are needed to make 500.0 mL of 5.0 M solution?

Given:
[tex]\[ \left( NH_4 \right)_2 SO_4 = 132.17 \, \text{g/mol} \][/tex]

[tex]\[ [?] \, \text{g} \][/tex]



Answer :

To determine how many grams of [tex]\(\left( NH _4\right)_2 SO _4\)[/tex] (ammonium sulfate) are needed to make 500.0 mL of a 5.0 M solution, follow these steps:

1. Convert the volume of the solution from milliliters to liters:
Since molarity ([tex]\(M\)[/tex]) is defined as moles of solute per liter of solution, we must work with the volume in liters.
[tex]\[ \text{Volume in liters} = \frac{\text{Volume in milliliters}}{1000} \][/tex]
For 500.0 mL:
[tex]\[ \text{Volume in liters} = \frac{500.0 \text{ mL}}{1000} = 0.5 \text{ L} \][/tex]

2. Calculate the number of moles of solute needed:
Molarity [tex]\(M\)[/tex] is defined as the number of moles of solute per liter of solution. Therefore, the number of moles needed can be found using the molarity and the volume of the solution in liters.
[tex]\[ \text{Moles of solute} = \text{Molarity} \times \text{Volume (in liters)} \][/tex]
Given that the molarity of the solution is 5.0 M and the volume is 0.5 L:
[tex]\[ \text{Moles of solute} = 5.0 \, \text{M} \times 0.5 \, \text{L} = 2.5 \text{ moles} \][/tex]

3. Calculate the mass of the solute needed:
To find the mass in grams of the solute ([tex]\(\left( NH _4\right)_2 SO _4\)[/tex]), use the relationship between the number of moles, the molar mass, and the mass.
[tex]\[ \text{Mass} = \text{Moles} \times \text{Molar mass} \][/tex]
The molar mass of ammonium sulfate is given as 132.17 g/mol and we need 2.5 moles:
[tex]\[ \text{Mass} = 2.5 \text{ moles} \times 132.17 \text{ g/mol} = 330.43 \text{ grams} \][/tex]

Therefore, the mass of [tex]\( \left( NH _4\right)_2 SO _4 \)[/tex] needed to make 500.0 mL of a 5.0 M solution is approximately [tex]\( 330.43 \)[/tex] grams.