# How would you prepare 200ml of a 4 M (NH_4)_2SO_4 (MW = 132.14 g / mol) ? Please show working

To solve for this problem we need to include our knowledge about molarity. Remember that Molarity is equal to the moles of solute per liter of solution. We can also write it as:

`Molarity = (mol es solute)/(Liter of solution)`

Since the question is asking how to prepare the solution,...

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To solve for this problem we need to include our knowledge about molarity. Remember that Molarity is equal to the moles of solute per liter of solution. We can also write it as:

`Molarity = (mol es solute)/(Liter of solution)`

Since the question is asking how to prepare the solution, we should  find what is the mass of `(NH_4)_2SO_4` . In solving that first we have to find the moles of solute.

`mol es of (NH_4)_2SO_4 = Molarity * Liters of solution`

`mol es of (NH_4)_2SO_4 = 4 (mol es)/(Liter) * (200)/(1000) Liters`

`mol es of (NH_4)_2SO_4 = 0.8 mol es`

**1000mL = 1L

Finally, we can get the mass of `(NH_4)_2SO_4` by multiplying the moles with the molar mass of `(NH_4)_2SO_4` .

`mass of(NH_4)_2SO_4 = (0.8 mol es)*132.14 (grams)/(mol)`

`mass of(NH_4)_2SO_4 = 105.712 grams`

To prepare the solution, we need to measure 106 grams of `(NH_4)_2SO_4` in a flask and fill it with water until the 200mL mark.