# What mass of solute is contained in a 675*mL volume of solution that was 0.025*mol*L^-1 with respect to "sodium sulfate"?

Apr 26, 2017

We get a mass of under $2.5 \cdot g$ of dissolved solute.

#### Explanation:

$\text{Concentration"="Moles of solute"/"Volume of solution}$, and thus, more symbolically, $C = \frac{n}{V}$.

As you progress in chemistry, you will use this relationship $\text{ad nauseum}$.

Given the formula:

$\text{moles"="concentration"xx"volume} .$ And thus..........

${n}_{N {a}_{2} S {O}_{4}} = 0.025 \cdot m o l \cdot \cancel{{L}^{-} 1} \times 675 \cdot \cancel{m L} \times {10}^{-} 3 \cancel{L \cdot m {L}^{-} 1}$

$= 0.016875 \cdot m o l$, note that dimensionally, the units we got were $\text{moles}$.

And for the mass of this quantity, we perform the product, $\text{mass"xx"molar mass}$ $=$ $0.016875 \cdot \cancel{m o l} \times 142.04 \cdot g \cdot \cancel{m o {l}^{-} 1}$
=??g