Re: molarity

In article <42e62feb$0$11352$3b214f66@xxxxxxxxxxxxxxxxxxx>,
"MMu" <brilhasti@xxxxxxx> wrote:
>
>"John Doe III" <john_doeIII@xxxxxxxxx> schrieb im Newsbeitrag
>news:D8nFe.8452$dU3.3134@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
>>
>> Let's assume 100 g of NaCl (MW = 58 g./ mol) is dissolved in 3.5 L of
>> water.
>> Is the molarity
>> A) 100 / (58 * 3.5)
>> or
>> becuase NaCl dissociates into 2 ions
>> B) 2 * 100 / (58 * 3.5)
>>
>> ???
>>
>> What is the molarity of a salt solution?
>>
>> is it the number of moles of all dissociated solutes that are dissolved in
>> the solvent divided by liter of solvent????
>>
>
>Ask yourself: what is molarity?
>Molarity is a number of "items" per litre.
>
>If you put a certain number of these items (Na and Cl atoms) in a container
>(your 3.5 liters) there still are that many in there.. water or not.
>
>If you ask the question of "what molarity does the NaCl solution have"
>you have to 'count' the number of "NaCl molecules" or rather, since those
>are dissolved and don't exist as such, the number of [(Na+) + (Cl-)] ions in
>water..
>
>[(Na+) + (Cl-)], Na+ ions, Cl- ions, all of these have the molarities of the
>equation under A.
>
>Think of it this way:
>Moles * [(Na+) + (Cl-)] = Moles*(Na+) + Moles*(Cl-)
>
>
>
Some books refer to the M times the number of ions as the "osmolarity", I
assume since osmotic pressure uses this quantity. But I always tell students
NEVER multiply the concentration times the number of particles; just multiply
the collig. prop. equation by that number (unless it's the v.p. of a salt
solution, in which case the calculation is different).
.

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