# Calculate molarity given density and mass percentage [closed]

### Problem

Calculate molarity of a solution of $$\ce{H2SO4}$$ with density $$\pu{1.198 g/cm3}$$ and containing $$27~\%$$ mass of $$\ce{H2SO4}.$$

$$\pu{3.374 M}$$

### My approach

I converted the density to g/l. Then i took 27% of the density and converted it into no of moles. Thus, came moles per litre, which is actually molarity. Yet, the answer key that my teacher had provided had a different answer compared to mine.

Is my method wrong or is there any calculation mistake?

• Write down the general algebraic calculation of molarity you have used. If it is right, result must be right as well. If you did trivial numeric error, it is easier to notice it this way The same if your algebra was wrong. – Poutnik Jul 5 at 7:22
• Sometimes a good strategy is to assume you have 1 g of solution (and therefore 0.27 g of acid), and work from there. Also, what is the molecular mass of H2SO4? – Buck Thorn Jul 5 at 7:49
• @BuckThorn I disagree with the usefulness of an approach where a physical quantity is fixed or its value taken arbitrary. When solving properly using algebraic equation, there is virtually no need to take any assumptions like that, which might be quite confusing and sometimes even wrong. P.S. also, on the nitpicking side, it's molar mass, not molecular mass you are probably referring to. – andselisk Jul 5 at 7:58
• @BuckThorn well, to me it's a useless concept I have never seen used anywhere else beyond introductory chemistry books of questionable quality that fellow physicists would relentlessly laugh at. A new variable is introduced out of nowhere that doesn't bring any new information, but causes confusion. "What if I only have 0.1 g of NaCl, how would I solve with 1 g?"; "how many sigfigs do I use with 1 g? Is 1.0000 g enough?"; "can I take 1 oz instead?"; "do I always plug in 1 <unit of measurement>?"; "can I plug in 1000 <unit of measurement> because there is "kilo"?" and so on. Just don't do it. – andselisk Jul 5 at 8:22
• @BuckThorn If it works for you, it's great. I would also use a mixing cross method occasionally in the lab and other not-exactly-correct things. But none of them should be forced as the primary method to the students. – andselisk Jul 5 at 8:33

### Solution

By definition, molarity $$c$$ is the amount of substance $$n$$ per volume of solution $$V$$:

$$c = \frac{n}{V}\label{eqn:1}\tag{1}$$

The volume $$V$$ can be found using total mass of solution $$m_\mathrm{tot}$$ and its density $$\rho$$:

$$V = \frac{m_\mathrm{tot}}{\rho}\label{eqn:2}\tag{2}$$

The amount of substance $$n$$ can be found using mass of solute $$m$$ and molar mass of solute $$M$$. Knowing mass fraction $$\omega,$$ one can switch to the mass of solution $$m_\mathrm{tot}$$ already used in \eqref{eqn:2}:

$$n = \frac{m}{M} = \frac{\omega m_\mathrm{tot}}{M}\label{eqn:3}\tag{3}$$

By plugging both \eqref{eqn:2} and \eqref{eqn:3} into \eqref{eqn:1}, one receives the final expression:

\require{cancel} \begin{align} c &= \frac{\omega \cancel{m_\mathrm{tot}}\rho}{M\cancel{m_\mathrm{tot}}} \\ &= \frac{\omega\rho}{M} \\ &= \frac{0.27\times\pu{1.198 g cm-3}}{\pu{98.08 g mol-1}} \\ &= \pu{3.30E-3 mol cm-3}~\text{or}~\pu{3.30 mol L-1}\tag{4} \end{align}

### Notes

• There is little to no point of converting anything prior to calculations as if the dimensional analysis is done right, you'll get the answer anyways.
• Density is an intensive property: you cannot use mass fraction to determine the density of solute alone.
• Regarding "no of moles": first, don't use non-standardized abbreviations like "no", and second, it's amount of substance, not "number of moles". "Number of moles" is an appropriate term in a dialogue with your neighbor about the population of talpids in your garden, but not in science.
• Don't forget to define each physical quantity properly and always try to solve the problem algebraically first, and plug in the values at the end when you have the ready-to-use equation for the quantity you were asked to find.
• Keep the track of significant figures (slang: "sigfigs"), e.g. you are given the density with four sigfigs, so it would make sense to also use molar mass with four sigfigs as well. Yet due to two sigfigs of mass fraction, the accuracy of the final answer is reduced.
• There is something to be observed that the answer is different from the given answer in my question. Thus, there can be two explanations, either my teacher made a mistake jn the answer key or u might've made a mistake. I am not sure but I too got the same answer as you – uddhav saikia Jul 6 at 15:06