# How do you calculate the moles and grams of solute in a solution of 1.0 L of 0.50 M NaCl?

May 13, 2017

$\text{Concentration"="Moles of solute"/"Volume of solution}$

#### Explanation:

And thus the quotient gives the units $m o l \cdot {L}^{-} 1$

We have a $0.50 \cdot m o l \cdot {L}^{-} 1$ solution of sodium chloride, and we use the product to give the molar quantity:

$\text{Volume"xx"Concentration} = 0.50 \cdot m o l \cdot \cancel{{L}^{-} 1} \times 1.0 \cdot \cancel{L} = 0.50 \cdot m o l$

$\text{Mass of solute} = 0.50 \cdot \cancel{m o l} \times 58.44 \cdot g \cdot \cancel{m o {l}^{-} 1} = 29.22 \cdot g .$

The inclusion of units like this in our calculations is an example of $\text{dimensional analysis}$. This is often helpful in calculations when you don't know first off whether you should divide or multiply. We wanted answers in (i) moles, and (ii) grams. The fact that dimensionally, the products gave us answers in units of grams and moles, persuades us that we got the order of operations correct.

For this reason, even tho it is admittedly a bit of a pfaff, I would routinely try to use units in my calculations. If I want an answer in $m o l \cdot {L}^{-} 1$, and get an answer with units ${\text{furlongs"*"fortnight}}^{-} 1$, I have probably done something wrong.