Question

What is the mass of `NaCl` required to prepare 0.5 liters of a 2.5 molar solution of `NaCl`?

Answer 1

73 g

Explanation:

The molecular weight of NaCl is 58.44 g/mol

So, one mole of NaCl weighs 58.44 g.

A 2.5 M solution is 2.5 moles per liter (Molarity is just the number of moles per liter).

Therefore, 0.5 L would contain 1.25 mol. Hence, you would need 1.25 × 58.44 g = 73 g.

As equations:

`M = "moles"/"vol"`

`M = "Molarity"`

`"vol" = "volume (in liters)"`

`"moles" = "grams"/(MW)`

`g = "weight of compound"`

`MW = "molecular weight"`

So,

`M = "moles"/"vol"`

Substitute for moles

`M = (g//MW)/"vol"`

We need `g`, so rearrange

`M × "vol" = g/(MW)`

`g = M × "vol" × MW`

Put in the numbers:

`g = "2.5 mol/L × 0.5 L × 58.44 g/mol"`

`g = "73 g"`

Personally, I prefer the first approach!

You can find out more about moles and molarity here.