Calculate the concentration in milimoles per litre (mM) of a solution that is 20% H2SO4 by mass and that has a density of 1.198 g/L 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!
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

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