- #1
EvilBunny
- 39
- 0
(purpose is to find molar mass)
At the start we dissolved 0.945 grams of unknown weak acid into water and made it up to
100 ml
Then we used 25 ml of that solution and titrated it until the end point.
Now , this is a monoprotic acid so it's going to take 1 mol of acid to neutralise 1 mol of base
22.1 ml of base was used
the conc of the base is 0.20 mol/l
So with this I can find the moles of acid and I got 4.42 x 10^-3
This is the part am unsure of , am I suppose to multiply that number of moles by 4 ?
because that was the number of moles there was in that 25 ml sample and we started it with 100 ml
If I do multiply by four I get 52
At the start we dissolved 0.945 grams of unknown weak acid into water and made it up to
100 ml
Then we used 25 ml of that solution and titrated it until the end point.
Now , this is a monoprotic acid so it's going to take 1 mol of acid to neutralise 1 mol of base
22.1 ml of base was used
the conc of the base is 0.20 mol/l
So with this I can find the moles of acid and I got 4.42 x 10^-3
This is the part am unsure of , am I suppose to multiply that number of moles by 4 ?
because that was the number of moles there was in that 25 ml sample and we started it with 100 ml
If I do multiply by four I get 52