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

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The concentration of a solution can be expressed in many way, one of them is molarity or the molar concentration. If a solution has a molarity of one, it has one mole of the solute in every liter of the solution.

Here we need to find the number of grams of ammonium sulfate needed to make 0.25 L of solution with a molarity of 6 M.

A solution with a molarity of 6 M has 6 moles of the solute per liter. 0.25 L of a solution with a molarity of 6M has 6*0.25 = 1.5 moles of the solute.

The molar mass of ammonium sulfate is 132.14 g/mole. The mass of 1.5 moles is 132.14*1.5 = 198.21 g.

Therefore 198.12 g of ammonium sulfate are required to make 0.25 L of a solution with a concentration of 6M.

**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 **

**M**olarity is defined as: ___________________________________

. liters of **solution**

. **[**g** (NH**4**)**2**SO**4**/**G.M.M**]**

**6 M = ______________________________**

. L **(NH**4**)**2**SO**4(aq)

. ** ? **g **/** 132g/mol

6 M = __________________

. 0.25 L

g **(NH**4**)**2**SO**4 = 6 mol/L * 0.25 L * 132 g/mol = 198 g**(NH**4**)**2**SO**4

Since **6M is only 1 significant figure** the answer to 1 significant figure would be 200 g **(NH**4**)**2**SO**4. 6.0M solution would also use **20**0 (with the bold numbers representing the significant figures)

If the solution were 6.00M (3 sig figs) then you would need 198 g**(NH**4**)**2**SO**4