# What is the concentration of a 10*cm^3 volume of a H_2O_4 at 9.8*mol*dm^-3 concentration that is diluted to a 100*cm^3 volume?

Mar 27, 2017

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

#### Explanation:

And given $\text{Concentration"="Moles of solute"/"Volume of solution}$, $\text{concentration}$ has units of $m o l \cdot {L}^{-} 1$, or $m o l \cdot {\mathrm{dm}}^{-} 3$.

Here we have (I think):

$\frac{10 \cdot c {m}^{3} \times {10}^{-} 3 \cdot {\mathrm{dm}}^{3} \cdot c {m}^{-} 3 \times 9.8 \cdot m o l \cdot {\mathrm{dm}}^{-} 3}{1000 \cdot c {m}^{3} \times {10}^{-} 3 \cdot {\mathrm{dm}}^{3} \cdot c {m}^{-} 3}$

$= 0.098 \cdot m o l \cdot {L}^{-} 1$ or $0.098 \cdot m o l \cdot {\mathrm{dm}}^{-} 3$, with respect to ${H}_{2} S {O}_{4} \left(a q\right)$.

This makes sense without all the pesky conversion factors, because you are diluting the original solution by a hundredfold.