Moles & Molar solutions.

[Gareth56]

If you want to make up for example a 1Molar HCl solution do you have to take into account the specific gravity of the conc. HCl that you're using? From memory I used to measure out 36.5mls of conc HCL and make that up to 1000ml in a volumetric flask. Is there a formula that you can use to tell you the amount of substance you need to make up a given molarity of solution?

 

It's straight forward if you're using solids because the formula if I recall is:-

 

No. of Moles required = weight in grams/molecular weight

 

e.g. a 0.5M solution of NaOH would be 20grams of NaOH made up to a 1000mls. I think:-)

 

Thanks

G56

[John Cuthber]

It's a bit more complicated than that.

If you took 36.5 ml of conc HCl it would weigh about 43 grams because the density is about 1.18 g/ml. Of that about 35% is actually HCl, the rest is water, so you would have 15g of HCl. That's only about 0.41 moles, so if you made it up to 1 litre you would have about 0.4M HCl.

[CharonY]

Also the density is dependent on the concentration of HCl 1.18 g/ml is at around 37%, around 32% it drops to roughly 1.15 g/ml

[Gareth56]

Thanks. Presumably the above applies to most or all the conc. acids like

H2SO4 and HNO3 and bases like NH3?

 

Also is there a difference between vol/vol & wt/vol solutions ?

 

I ask because do you have to take the SG into account if your making up a solution in vol/vol or only if you're using the wt/vol method?