[trinhn812]

Values of the weight percentage of the acid and the density (at 20°C of concentrated aqueous solutions of reagent grade acids commonly provided by chemical manufacturers are:
      HCl   36.0%   1.19 g/cm3

Calculate the molarity and molality of acid in each of these solutions.  What mass of each of these solutions must be taken to prepare 1.00 L of solution with a concentration of 0.100 M?


So I figured out molarity by doing:

1.19g/mol * 1000mL/L = 1190 g/L
36% = (x/1190g) * 100%
428.4 = x = g of HCl

1190g-428.4g = 761.6g H2O

428.4gHCl*molar mass of HCl =11.75 mol

11.75 mol / .761kgH20 = 16.4 mol/kg


Now I don't know how to find the molarity and I could use some hints for the second part of the problem that asks you for the the mass of each solution.

Please help and thanks.

[LQ43]

1. remember the equation for Molarity = mol / L

you calculated mol HCl  11.75 mol (think you meant 428.4 / MM HCl) and this is in 1L of solution

so M = mol / L of solution =

2. Molality is okay

                 

3. for dilution use Mconc x V conc = Mdil x V dil    to find Vconc,

then change to g using the density

[Alpha-Omega]

This may help: And BTW do not feel bad you have no idea how many people cannot do this calculation....and how many forget to use the density when usin acids like HCl and H2SO4..

Molarity:  Moles/L

HCl = 36.0%
Density = 1.19 g/cm^3  = 1.19 g/mL
GFW HCl = 36.45 g/mol

To find the molarity M use this: M = mol/L…those are you units

Use the density to calculate the weight of 1 liter of solution-
1000 mL x 1.19 g/mL = 1190 grams
Use the percent to calculate the mass of HCl in 1190 g (1.000 liters)-
1190 x  36.0% =  1190 x 0.36 = 428.4 g HCl


Now you know the weight of HCl per liter.

Convert the weight of HCl to moles of HCl-
428.4 g / 36.45 g/mol = 11.75 moles of HCl in 1.000 liter
Molarity = 11.75 M

[Alpha-Omega]

To find the mass from here is very simple:

Use M1V1 = M2V2

Or as previously stated:  Mconc  x Vconc = Mdil x Vdil

Rearranging gives you Vconc = (Mdil x Vdil)/Mconc

Now you have your volume of concentrated HCL

Convert that to mL and the multiply by the density to get rid of the mL and report your answer in g.

[Borek]

http://www.chembuddy.com/?left=concentration&right=percentage-to-molarity

http://www.chembuddy.com/?left=concentration&right=dilution-mixing

[ARGOS++]

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 MolaRity is just  11.75!
(Remember: MolaRity = moles per Liter final Solution!)

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

So the MolaRity of your reagent grad HCl is 11.75, and
     the MolaLity of the same Solution is      15.42.

As the Question asks for the required mass to make a 0.1 molaR 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⁺⁺