How to Work Out moles of NaOH 0.10 mol L^-1 if 25 mls used. (1 Viewer)

[Lordsion]

How do you work out the moles if (a) concentration is 0.10 mol L^-1 and (b) you use 25 ml


Moles = M/MM

So how can i work it out with just that info?

EDIT:

c = moles / V (L)

0.10 = ? / 0.0025
? = 0.0025 X 0.10

moles = 0.00025 ? correct?


and say for vinegar of concentration 0.119 mol L^-1 .... and i used 21 mls ...... this would mean:

? = 0.0021 X 0.119
moles = 0.0002499

 

[vds700]

your working is incorrect

25 mL = 0.025 L

[NaOH]= n/V
0.10 = n/0.025
n = 0.025 x 0.10
=0.0025 mol NaOH

 

[Lordsion]

How do you work out the moles if (a) concentration is 0.10 mol L^-1 and (b) you use 25 ml


Moles = M/MM

So how can i work it out with just that info?

EDIT:

c = moles / V (L)

0.10 = ? / 0.0025
? = 0.0025 X 0.10

moles = 0.00025 ? correct?


and say for vinegar of concentration 0.119 mol L^-1 .... and i used 21 mls ...... this would mean:

? = 0.0021 X 0.119
moles = 0.0002499

~~~~

And also, with the vinegar i worked out the concentration of the vinegar with the volume coming from the titre (amount expensed from burette in tritration) --- is this the right value to use for the volume in the equation above ( moles = C x V) to figure out moles of vinegar?

 

[vds700]

21 mL is equal to 0.021 L not 0.0021 L as in your calculation. To convert from millilitres to litres, you simply divide by 1000. There are 1000 mL in 1 L.

I'm assuming that NaOH is your standard solution, 25 mL in a conical flask and you added 21 mL of an unknown concentration of vinegar from your burette. correct me if this is wrong.

The chemical equation for this reaction shows a 1:1 molar ratio between reactants and products. ie, 1 mol of NaOH will neutralise 1 mol of vinegar.

You have 0.0025 mol NaOH therefore you need 0.0025 mol vinegar to completely neutralise the base.

[vinegar]= n/V
=0.0025/0.021
=0.119 mol/L