Calculating the Concentration of N/5 HCl After Removal of 3.65g

[xiphoid]

Homework Statement

On boiling 1litre of N/5 HCl, the volume of the aqueous solution decreases to 250ml. If 3.65g of HCl is removed from solution, then the concentration of resulting solution becomes?
N/5
N/2.5
N/20
N/10

Homework Equations

HCl=36.5g/ mol

The Attempt at a Solution

no idea

 

[Borek]

Use concentration definition. You start with 1L of solution - how much solvent? How many moles of HCl? How much solvent after the boiling? How much HCl?

 

[ohms law]

At this point you should know about Molality.
There should be an equation for it (Molality) in your textbook... I think that Borek wants you to look it up rather than me tell you what the equation is. You'll really need to memorize what the equation is anyway (don't worry, it's straight forward).

 

[JohnRC]

At this point you should know about Molality.
There should be an equation for it (Molality) in your textbook... I think that Borek wants you to look it up rather than me tell you what the equation is. You'll really need to memorize what the equation is anyway (don't worry, it's straight forward).

not molality (m) but molarity (M). There is a subtle, but important difference. And I do not think that that is where your problem lies because normality (N) is a "drop in" substitute for molarity (M) for monofunctional reagents.

"Equivalent weights" and "Normality" are no longer used by chemists (since a IUPAC ruling in the mid 1960s), but they have lingered on in engineering and in sections of the chemical industry. They become problematic because of ambiguities, where, for example, nitric acid as oxidant may have functionalities of 1, 3, 4, or 5, for its various oxidation reactions, meaning that the exact same solution could be regarded as 0.1 N, 0.3 N, 0.4 N, or 0.5 N!

 

[ohms law]

oops, sorry. That's what I meant (Molarity). Damn.
I've never heard of "normality" before.

Molality (m=mol/kg) is what we were doing about two weeks ago, so... *shrug*