Mole equations!?

Question

Hi im currently doing an assignment on a CSI. I carried out a titration but im stuck on working out the molar equations.

I had a bench acid of HCl at 2mol, but i diluted 1cm3 of it with 9 cm3 of distilled water.

How do i work out how many moles were used?

Answer 1

typically you do not assume conservation of volume. ie you do not assume 1 cm3 of (water + HCl) + 9 cm3 water = 10 cm3. You would normally put the water/HCl in a volumetrc flask and dilute to the line to correct for this.

that said, in your case, I suggest you assume conservation of volume and say...

Also, I going to assume you mean the HCl is 2 molar. Is that correct?

C1V1 = C2V2
C1 = 2 moles/liter
V1 = 1 cm3 = 1 ml
V2 = 1 cm3 + 9 cm3 = 10 cm3
so c2 = 2 moles/liter x 1 ml / 10 ml = 0.2 moles / liter

so then if you used "V" milliters to titrate to endpoint then the moles used would be 0.2 moles/liter x V ml x (1 liter / 1000 ml) = ...

Answer 2

HCl @2mol.
This means 2 moles per litre (mol dm^3)
One litre contains 1000 cm^3
so in 1 cm3 you have 2/1000 = 0.002 mols.
If you added 9 cm3 there is now 10cm3 containing 0.002 mol.
The new conc'n is 10cm3 x 100 = 1000cm3
0.002 x 100 = 0..2 mol dm^-3

However, is only 1 cm^3 was used then only 0.002 mols is used
and adding 9 cm^3 there is still only 0.002 mols used but it is diluted. to a conc'n of 0.2 mol dm^-3

Answer 3

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Answer 4

ok if you used 2M HCl Acid and you have 1 cubic centimetre of it, then assuming that it has the density of water, you began with 0.002 mols of HCl. After adding the 9cm 3 of water the new concentration of your acid is 0.2M

Just remember n=cV

Answer 5

Read the text books like I did!!

Answer 6

there are three equations to calculate number of moles
n=mass/molarmass
n=volume/24 (in dm3) or n=volume/24000 (in cm3) (ONLY FOR GASES)
n=concentration * volume (ONLY FOR LIQUIDS)

Answer 7

Always remember----equal vols equal mols.