# How do you calculate the moles and grams of solute in a solution of 5.0 x 10^2 mL of 2 M KNO_3?

May 11, 2017

There are $1 \cdot m o l$, and approx. $100 \cdot g$ of nitrate.......

#### Explanation:

We use the relationship:

$\text{Molarity"="Moles of solute"/"Volume of solution}$, and thus to get the number of $\text{moles}$ we multiply the $\text{volume}$ by the $\text{molarity}$ to get an answer in $\text{moles} .$

And for your problem, there are:

$5 \times {10}^{2} \cdot m L \times {10}^{-} 3 \cdot m L \cdot {L}^{-} 1 \times 2.0 \cdot m o l \cdot {L}^{-} 1 = 1 \cdot m o l$ with respect to $K N {O}_{3}$.

Note the use of the units in the calculation. We wanted an answer in $m o l$; the calculation gave us an answer in $\text{mol}$. Chemists (and physical scientists) often include units in their calculations to ensure that they perform the right order of operations: $\text{do I divide or do I multiply?}$ Anybody can make a mistake, and everybody has.

And since $\text{moles"="mass of solute"/"molar mass of solute}$, there are $1 \cdot \cancel{m o l} \times 101.10 \cdot g \cdot \cancel{m o {l}^{-} 1} = 101.1 \cdot g$