Dear **Trinhn812**;

Don’t get confused of so much different (and sometimes ?difficult?) explanations and start just with your last correct calculated result: 11.75 moles HCl!

As you explained that this 11.75 moles are contained in 1.00 Liter (= 1.19 kg) of **final Solution**, so the Mola**R**ity is just 11.75!

(**Remember:** Mola**R**ity = moles per **Liter** **final Solution**!)

But as Mola**L**ity is defined as moles per **Kilogram** **Solvent only** your Mola**L**ity will be = Mola**R**ity divided by mass of solvent = 11.75 (moles/Liter) / (1.19 - 0.482) (kg/Liter) = 15.42 (moles/kg Solvent)

So the Mola**R**ity of your reagent grad HCl is 11.75, and

the Mola**L**ity of the same Solution is 15.42.

As the Question asks for the required **mass** to make a 0.1 mola**R** Solution you can simply calculate from your first result 1.19 kg / 11.75 moles * 0.1 moles = 0.01013 kg

**Conclusion:** Weight 10.13 g of your concentrated HCl into a volumetric flask of 1.0 Liter and fill with (dest.) water till to the mark! That makes 1.0 Liter with 0.10 moles HCl.

And that’s just all.

Good Luck!

ARGOS^{++}