0
$\begingroup$

Calculate the concentration in milimoles per litre ($\pu{mM}$) of a solution that is $\pu{20\%} \ce{H2SO4}$ by mass and that has a density of $\pu{1.198 g/L}$.

For the answer, please provide two decimal places.

I calculated an answer but I'm just checking to see if I am correct...

I got $\pu{61.7 mM}$

First, I calculated the molar mass of $\ce{H2SO4}$ which is $\pu{98.086g/mol}$ and then I multiplied it by $\pu{20\%}$ which then got me $\pu{19.6172g/mol}$.

Second, I divided density by mass to cancel out the grams to get $\pu{mol/L}$ which then came out to $\pu{0.06106M}$ and then I divided it by $1000$ to get $\pu{mM} = \pu{61.07mM}$

If someone could check if this is correct that would be great, thanks!

$\endgroup$
1
$\begingroup$

the solution has a density of 1.198 g/L.

This has to be wrong. I assume that the density is 1.198 g/ml.

Thus a liter has a mass of

$$1000\text{ ml/L} \times 1.198\text{ g/ml}= 1198\text{ g/L}$$

20% of the mass is $\ce{H2SO4}$, thus the mass of $\ce{H2SO4}$ in 1.000 L of solutiom is:

$$0.20 \times 1198\text{ g/L} = 239.6 \text{ g/L}$$

$\ce{H2SO4}$ has a molecular mass of $98.079\text{ g/mol}$ so the number of moles is

$$\dfrac{239.6 \text{ g/L}}{98.079\text{ g/mol}} = 2.443\text{ moles/L}$$

Now converting to millimoles

$$2.443\text{ moles/L}\times 1000\text{ mM/mole} = 2443\text{ mM/L} \ce{->[rounding]} 2.4\times 10^3\text{ mM/L} $$

$\endgroup$
0
$\begingroup$

Always best to write these out with unit conversions:

$$\frac {1\ \pu{gram}\ \ce{[H2SO4]}}{\pu{gram}\; [solution]} \mathrm{x} \frac {1.198\; \pu{gram}\; [solution]} {1\; \pu L\ [solution]} \mathrm{x} \frac{1\; \pu{mol}\; [\ce{H2SO4}]}{98.086\ \pu{gram}\ [\ce{H2SO4}]} \mathrm{x} \frac {1000\ \pu{mmol}}{1\pu{mol}}$$


I encourage you to write this one out to check your answer! It looks like you have flipped a fraction somewhere.

$\endgroup$
  • $\begingroup$ Welcome to Chemistry.SE! Good effort, but please note the edits and how much more simple and clean the post is. Please note that formulas can be better expressed with \$\ce{}\$ for chemical formulas/equations, \$\mathrm{}\$ for math term/equations, and \$\pu\$ for units. More information is available in this meta post Also, take a minute to look over the help center and tour page to better understand our guidelines and question policies. $\endgroup$ – A.K. Sep 18 '18 at 4:21

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.