Molality calculation.... Is it correct?

[sp3sp2sp]

Homework Statement

Commercial grade HCl solutions are typically 39.0% (by mass) HCl in water. Determine the molality of this HCl solution, if it has a density of 1.20 g/mL.

Homework Equations

molality = moles solute / kg solvent

The Attempt at a Solution

39.0g HCl (1mol/36.46g/mol) = 1.07mol HCl

100g solution (1mL/1.20g) (1L/1000mL) = 0.0833L

1.07mol HCl / 61.0g H2O = 0.0175m = molality

thanks for any help

 

[Charles Link]

It seems they are supplying what may be two different criterion of what the solution consists of. Is it 39% HCl by mass or is it 1.20 g/ml ? For the second number, you need to know the density of liquid HCl and assume its volume is unchanged when mixed with water. At least that's what I would try.
@Borek I welcome your input here, because I'm not sure I got it right. $\\$ Edit: A google shows HCl is a colorless gas at room temperature, so the liquid idea that I have above will not work. $\\$ The statement of the problem isn't clear, but I think you can assume the 1.20 gram is 39% HCl and 61% water. From there, it should be a simple calculation. You need to determine how much HCl you have in grams per kilogram of water, and then convert to moles. And I'm pretty sure this second approach is the correct one=the problem is fairly simple.

 

[Borek]

1.2 g/mL is definitely the density of the solution, so yes, the most obvious approach is to follow path: volume (assume 1 L for ease of calculations) → mass of the solution → mass of HCl in that solution → number of moles → concentration.

 

[mjc123]

For molality (as distinct from molarity) the density is irrelevant. 39% by mass HCl is 39g HCl (1.07 mol) to 61g water, whatever the volume. OP's only mistake was to calculate moles per g water rather than moles per kg.

 

[Borek]

Oh my, you are perfectly right, missed that "l" :(

 

[Charles Link]

That's why this problem is somewhat unclear. If the density is indeed 1.20 g/ml, it could mean that the molality of the solution could be far different than what is computed assuming 39% HCl by mass. If it is 39% by mass, you get one answer. $\\$ There is insufficient information to do much with the 1.20 g/ml number. When they say "typically, they don't really state precisely that the solution at hand has that 39% by mass concentration. $\\$ Looks like from this data set, the percentage HCl by mass is indeed about 40% (close to 39%) when the density is 1.20 g/ml : http://www.starch.dk/isi/tables/hcl.htm If they gave you the table, along with the 1.20 g/ml number, you could conclude a 40% by mass composition and compute the molality.

 

[sp3sp2sp]

This problem is confusing me..or its too easy and I am missing something dumb. Isnt the density of solution value sort of pointless in this problem?
molality = moles solute / kg solvent
So to get moles of HCl : 39.0g HCl (1mol/36.46g/mol) = 1.07mol HCl
And to get kg of solvent , which is water is just 100g of solution - 39g HCl = 61g of water *(1kg/1000g) = .061kg

Then divide 1.07mol/.061kg = 17.54m of solution
Is that correct? thanks

 

[Borek]

Yes, 17.5 m looks OK.

 

[sp3sp2sp]

thanks ...so I guess the density of solution is like a diversion or something (not relevent)

 

[Charles Link]

thanks ...so I guess the density of solution is like a diversion or something (not relevent)

See my post 6, with the "link" to the table. If they provided you with that table, instead of the 39% number, you could have solved the problem with the look-up table by seeing that 1.20 g/ml corresponds to 40% HCl by mass $\approx$ 39%
.

 

[sp3sp2sp]

oh i didnt know there's tables for this ..but there was no other info, not even part of a lab and not connected to another question or anything

 

[Charles Link]

oh i didnt know there's tables for this ..but there was no other info, not even part of a lab and not connected to another question or anything

I got lucky with a google. In general, as a question on a test, there was insufficient information provided to solve it using the 1.20 number. The 39% number was sufficient in itself though.

 

[Borek]

Actually I think now that fact that both % and density were given was what made me miss the fact it asks about molality, as these are information necessary or molarity calculation. Routine killed attention.