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

Aug 17, 2016

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:

$M = \text{moles"/"vol}$

$M = \text{Molarity}$

$\text{vol" = "volume (in liters)}$

$\frac{\text{moles" = "grams}}{M W}$

$g = \text{weight of compound}$

$M W = \text{molecular weight}$

So,

$M = \text{moles"/"vol}$

Substitute for moles

$M = \frac{g / M W}{\text{vol}}$

We need $g$, so rearrange

M × "vol" = g/(MW)

g = M × "vol" × MW

Put in the numbers:

$g = \text{2.5 mol/L × 0.5 L × 58.44 g/mol}$

$g = \text{73 g}$

Personally, I prefer the first approach!

You can find out more about moles and molarity here.