# Find molar concentration of sodium carbonate dissolved in water

The question is

A solution of $\ce{Na2CO3}$ is made by dissolving $5.3~\mathrm{g}$ of the powder in enough water to make $250~\mathrm{mL}$ of solution. Find the molar concentration.

I have done the following but I don't know why it's not working.

Molar mass: $106~\mathrm{g}$

$5.3~\mathrm{g} / 106 = 0.05\ \mathrm{mol}$

$0.05\ \mathrm{mol} / 250\ \mathrm{mL} = 0.0002\ \mathrm{mL}$

However, the answer is $0.2\ \mathrm{mol/L}$ but I got $0.0002$. It does become $0.2$ when I multiply by $1000$ but the answer is already in $\mathrm{mL}$. How do I arrive at the correct answer?

• Please fix your units! – Jan Jun 1 '16 at 22:26

# Use units! Consistently, properly, all the time!

You have units all over the place sometimes fitting the quantity you are calculating, most often not. If you approached the problem with correct use of units, there would be no question.

First, the molar mass.

$$M(\ce{Na2CO3}) = 105.99~\mathbf{\frac{g}{mol}}$$

Grams are not equal to grams per mole. Then, let’s do the rest without losing or inventing units as we go along. (I bolded the units for you.)

$$n(\ce{Na2CO3}) = \frac{m(\ce{Na2CO3})}{M(\ce{Na2CO3})} = \frac{5.3~\mathbf{g}}{105.99~\mathbf{\frac{g}{mol}}} = 0.05~\mathbf{mol}$$

$$c(\ce{Na2CO3}) = \frac{n(\ce{Na2CO3})}{V} = \frac{0.05~\mathbf{mol}}{250~\mathbf{ml}} = \frac{0.05~\mathbf{mol}}{0.250~\mathbf{l}} = 0.2~\mathbf{\frac{mol}{l}}$$

Different versions of jumbling around the powers of 10 (especially in the millilitre part) will get you the same result if done correctly. It really bears repeating.