Please help?

I hate to ask a retardedly simple question, but for some reason I cannot currently comprehend this problem.

A solution of 38% HCl by mass has a density of 1.19g/cm^3.
if 3.65kg of HCl is needed, what volume in liters, of HCl solution is required?

The answer is 3.07L, but I don't understand how to figure that out.

Actually, I get 8.1 L.

First, let's convert the density of the solution to something we can work with.

1 mL = 1 cm^3
1 L = 1000 cm^3

1.19 g/cm^3 = 1.19 g/mL = 1.19 kg/L

Next, let's calculate how much HCl is actually in that liter of solution.

38% of 1.19 kg/L -> 0.38 * 1.19 kg/L = 0.4522 -> (significant figures) -> 0.45 kg

So we see that in each liter of solution (1.19 kg), there is 0.45 kg of HCl.

Next, we calculate how many liters we need to get 3.65 kg of HCl, if each liter gives us 0.45 kg.

3.65 kg / 0.45 kg = 8.111r1 -> (significant figures) -> 8.1 -> 8.1 L

So 8.1 L of solution is needed to have 3.65 kg of HCL.

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CHECKING OUR WORK
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The solution has a density of 1.19 kg/L. We have 8.1 L of solution.

8.1 L * 1.19 kg/L = 9.639 kg -> (significant figures) -> 9.6 kg

So we have 9.6 kg of solution. The solution is 38% HCl by mass.

9.6 kg * 0.38 = 3.648 kg of HCl -> (significant figures) -> 3.7 kg of HCl.

o hai