molarity vs concentration

wolf99

Hi

Is molarity the same as concentration????

If somebody asks me what concentration electrolyte I need and I reply
"1.75 moles per liter" will that be what they expected, and is this enough information for them to make up the solution for me????

thanks

cianl1

Hi Wolf

Yes, molarity is the same as concentration. It's a measure of the no. of moles of a substance in a certain volume of solvent.

Taking your example above 1.75 moles per litre (mol/L) is 1.75 moles of, let's say NaOH, in a litre of water.

Any lab technician worth their salt would be able to make up a solution of the concentration by simply giving them that information.

wolf99

OK brilliant thanks.

however if I have to make a solution up myself. I have some de-ionised water.
and hopefully by the time I get to actually doing this I'll have some high grade sodium chloride.
My calculations tell me I need 1.095 mol/L. I assume I dont just carve up
1.09-ish crystals of salt.

How does one "weigh" moles? or is a mole equal to a gram or something?

thanks again, I am so not a scientist, so trying to understand by looking at google info on this stuff just escapes me!

cianl1

wolf99 wrote: »

OK brilliant thanks.

however if I have to make a solution up myself. I have some de-ionised water.
and hopefully by the time I get to actually doing this I'll have some high grade sodium chloride.
My calculations tell me I need 1.095 mol/L. I assume I dont just carve up
1.09-ish crystals of salt.

How does one "weigh" moles? or is a mole equal to a gram or something?

thanks again, I am so not a scientist, so trying to understand by looking at google info on this stuff just escapes me!

Ok, you need to make a 1.095 mol/L solution.

We use the equations:

n = m/Mr and M = n x 1000/V

Where n = moles
m = mass
Mr = molar mass
M = concentration
and V = volume

First rearrange the second equation so it reads:

n = (M x V)/1000, this will give you the number of moles required.

You have the concentration (i.e. 1.095 mol/L or 1.095 M) and the volume is the volume of solution made up, for the purpose of the example we'll say it's 50 cm^3 (cm cubed)

So, n = (1.095 M x 50 cm^3)/1000

which worked out gives n = 0.05475 moles

Now rearranging our other equation:

m = Mr x n

The molar mass of sodium chloride is given by adding their mass numbers.
(Na = 22.9 Cl = 35.5 => 22.9 + 35.5 = 58.4)

m = (58.4 g/mol)(0.05475 mol)

Multiply it out and you have the mass of NaCl required:

m = 3.19 g

wolf99

Hey Cian,

thanks for sticking with it! thats a lot clearer than what Ive been struggling with

muchos gracias.

wolf99

Hey
So I got round to adding this to my excel sheet, And i'd like to double check if I've got it all right.
(we all love to be right, right?)

I start out with required conductivity of: 10%,

convert to parts per million:
(10%) x (0.64) = 6.4ppm,

calculate moles per litre of NaCL:
(6.4ppm) / (58.443g/mol) = 0.1095mol/L, (this is 1/10th of what I previously mentioned to Cian, so the answer should be reduced by an order of 10)

use to calculate moles:
((0.1095mol/L) x 50cm^3)) / 1000 = 0.00547moles

and so finally mass:
(58.443g/mol) x (0.00547moles) = 0.32grams (exactly 1 order of 10 less)

Is this the correct method?
Thanks for bearing with me!

EDIT:
hmm just noticed this gives me 0.32g for every solute weight (g/mol) I try even though those weights are different...

That cant be right??

wolf99

For example, HF

weight: 20.00634

conductivity again at 10% to ppm: 10 * 0.64 = 6.4ppm

molarity: 6.4 / 20.00634 = 0.3198

moles: (0.3198 * 50cm^3) / 1000 = 0.1599

mass: 20.00634 * 0.1599 = 0.32g !!!

cianl1

You got your moles arseways there.

(0.3198 * 50cm^3)/1000 = 0.01599

Otherwise, it checks out.

What I would suggest is look again at your concentration since ppm is not the same as g/L.

wolf99

oops I should have said 100uS/cm not 1005 or 10% thats were I got mixed up...

still leaves me with 3.2g for all solutes....