# "Solute 1", "S1", has a molecular mass of 40.0*g*mol^-1...what is...?

## ...what is $\left[S 1\right]$ if a $4 \cdot g$ mass of this material is dissolved in a $200 \cdot m L$ volume of solution?

Aug 29, 2017

By definition, $\text{concentration"-="amount of stuff"/"volume of solution}$

#### Explanation:

And typically, we use $\text{molarity"-="moles of solute"/"volume of solution}$

And thus $\text{concentration of S1}$

$= \frac{\frac{4 \cdot g}{40.0 \cdot g \cdot m o {l}^{-} 1}}{200 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1} = 0.50 \cdot m o l \cdot {L}^{-} 1$

And note the dimensionality of this solution....

We cancel out common terms in the numerator and denominator....

$= \frac{\frac{4 \cdot \cancel{g}}{40.0 \cdot \cancel{g} \cdot m o {l}^{-} 1}}{200 \cdot \cancel{m L} \times {10}^{-} 3 \cdot L \cdot \cancel{m {L}^{-} 1}} = 0.5 \cdot \frac{\frac{1}{m o {l}^{-} 1}}{L} = 0.50 \cdot m o l \cdot {L}^{-} 1$

Why? Because $\frac{1}{x} ^ - 1 = \frac{1}{\frac{1}{x}} = x$ as required................