Need Help With Chemistry Calculations

[satty2007]

I used moles=mass/Mr =1.486x10^-4 moles

c=nv/1000 = (1.486x10^-4 x 250)/1000 = 3.72x10^-5 M (37uM)

If you take 5mls or 10mls of this solution now, it should remain at the same concentration, as you have not diluted it nor concentrated it. The number of moles in that solution will have changed though.

Yh thanks i get that, one more question, if you take out 5ml or 10 ml then top it up with more nitric aicd in a 25 ml volumetric flask, wouldnt the concentration decrease?????

[I humbly bow]

I used moles=mass/Mr =1.486x10^-4 moles

c=nv/1000 = (1.486x10^-4 x 250)/1000 = 3.72x10^-5 M (37uM)

If you take 5mls or 10mls of this solution now, it should remain at the same concentration, as you have not diluted it nor concentrated it. The number of moles in that solution will have changed though.

your second equation is wrong c = n/v, do it should be c = (n/v)/1000 so c= 5.945 x 10-4

5ml in 25 = (5.945 x 10-4) / 5

10ml in 25 = (5.945 x 10-4) / 2.5

i got asked a similar qu. to this in my interview, when i first got my job, they didn't correct me...so i'm assuming that's right lol!

"Yh thanks i get that, one more question, if you take out 5ml or 10 ml then top it up with more nitric aicd in a 25 ml volumetric flask, wouldnt the concentration decrease?????"

you mean the original 250ml flask? if you topped up with 15ml more nitric acid, to replace what you took out? yeah that would decrease the concentration.... 15/235 = 0.0638

so 0.0638 x (5.945 x 10-4) gives your new concentration.

[Wicked Warrior]

I used moles=mass/Mr =1.486x10^-4 moles

c=nv/1000 = (1.486x10^-4 x 250)/1000 = 3.72x10^-5 M (37uM)

If you take 5mls or 10mls of this solution now, it should remain at the same concentration, as you have not diluted it nor concentrated it. The number of moles in that solution will have changed though.

your second equation is wrong c = n/v, do it should be c = (n/v)/1000 so c= 5.945 x 10-4

5ml in 25 = (5.945 x 10-4) / 5

10ml in 25 = (5.945 x 10-4) / 2.5

i got asked a similar qu. to this in my interview, when i first got my job, they didn't correct me...so i'm assuming that's right lol!

"Yh thanks i get that, one more question, if you take out 5ml or 10 ml then top it up with more nitric aicd in a 25 ml volumetric flask, wouldnt the concentration decrease?????"

you mean the original 250ml flask? if you topped up with 15ml more nitric acid, to replace what you took out? yeah that would decrease the concentration.... 15/235 = 0.0638

so 0.0638 x (5.945 x 10-4) gives your new concentration.

Oversight.

n=cv/1000 --> c=1000n/v

Sorry bout mistake.

[satty2007]

"you mean the original 250ml flask? if you topped up with 15ml more nitric acid, to replace what you took out? yeah that would decrease the concentration.... 15/235 = 0.0638"

No i was askin the 5 ml and 10ml taken out from the 250 solution, and put in to 25ml volumetic flask and topped up with nitric aicd, would they have a lower concn then the 250ml or will it be the same???

by the way thanks for your help :TH:

[I humbly bow]

"you mean the original 250ml flask? if you topped up with 15ml more nitric acid, to replace what you took out? yeah that would decrease the concentration.... 15/235 = 0.0638"

No i was askin the 5 ml and 10ml taken out from the 250 solution, and put in to 25ml volumetic flask and topped up with nitric aicd, would they have a lower concn then the 250ml or will it be the same???

by the way thanks for your help :TH:

yeah they will be less, your diluting them....the answer is above;

5ml in 25 = (5.945 x 10-4) / 5

10ml in 25 = (5.945 x 10-4) / 2.5

hope that helps