In the the formula

$$C = 10P\frac{D}{M}$$

$C$ is the concentration of a substance in a solution, $P$ is the percent concentration (purity) of the substance, $D$ is the relative density of the solution and $M$ is the molar mass of the substance.

Say $P = 60\%$, do we substitute $60$ or $60/100$ in the formula? The proofs I've read suggest $60$.


It all depends on the units other variables are expressed in. By dimensional analysis it's obvious that you don't expect percent to appear in the final result, so $P$ is expressed as a fraction. Using percentage number can be justified if $100\%$ is conveniently cancelled out.

The $10$-multiplier suggests that this might be case here, e.g. if $D$ is expressed in $\pu{g cm-3}$, $M$ — in $\pu{g mol-1}$, then in order to obtain $C$ in $\pu{mol dm-3}$ (or $\pu{mol L-1}$, which is the same), you must plug in the value for $P$ in percents:

$$[C] = \frac{\pu{g cm-3}}{\pu{g mol-1}} = 10^3\cdot\pu{g dm-3} = 10\cdot [P~\text{in}~\%]\cdot\pu{g dm-3}$$

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.