Need Help With Chemistry Calculations

[satty2007]

Fateh

Im really stuck on my chemsitry calculations, could some one please help me out. Its to do with dilution factors.

I made solution of a sample of Nd2O3, molecular mass = 336.4

I used 0.05g and addded it to 250ml of water, so the no of moles of Nd in 250ml is 1.49x10-4 and so the concentration in 1 litre would be 2x (5.96x10-4moldm-1) as there are 2 moles of Nd.

I then took 5 ml and 10 ml out of the 250ml flask and put it in a 25ml flask reapectivley.......i wanted to know how would you calulculate the concentration of the 5ml and 10ml solutions???? :gg:

Thanks in advance

[Guest peacemaker]

Try

www.Physicsforums.com

[Revolutionary]

I used 0.05g and addded it to 250ml of water, so the no of moles of Nd in 250ml is 1.49x10-4 and so the concentration in 1 litre would be 2x (5.96x10-4moldm-1) as there are 2 moles of Nd.

btw dont use rounded figures when mulitplying, its actually 5.95 not 5.96..

OK so the answer u reach is:

0.001189 (4sf) moles of Nd in every L of water right?

s'all good so far

I then took 5 ml and 10 ml out of the 250ml flask and put it in a 25ml flask reapectivley.......i wanted to know how would you calulculate the concentration of the 5ml and 10ml solutions????

Concetration x Volume = No. of mols

With the info you got: Concentration x 0.05L = 0.001189 (4sf) mol

Therefore Concentration = 0.02378 mol/L

BUT BUT BUT BUT BUT BUT BUT BUT BUT BUT!!!!

having said that: If you take out 5ml of a concentrated solution (lets say 1 unit) then the concentration will remain the same.. no? Maybe the question you asked is wrong.. infact the concetration will remain the same no matter how much of the solution you will take out!

[Guest _guest_]

Sorry,

Just a silly question what is Nd. is it from heavy elements f series?

I assume a hypothetical element or is it naming convention of some solution.

Regards

[satty2007]

I used 0.05g and addded it to 250ml of water, so the no of moles of Nd in 250ml is 1.49x10-4 and so the concentration in 1 litre would be 2x (5.96x10-4moldm-1) as there are 2 moles of Nd.

btw dont use rounded figures when mulitplying, its actually 5.95 not 5.96..

OK so the answer u reach is:

0.001189 (4sf) moles of Nd in every L of water right?

s'all good so far

I then took 5 ml and 10 ml out of the 250ml flask and put it in a 25ml flask reapectivley.......i wanted to know how would you calulculate the concentration of the 5ml and 10ml solutions????

Concetration x Volume = No. of mols

With the info you got: Concentration x 0.05L = 0.001189 (4sf) mol

Therefore Concentration = 0.02378 mol/L

BUT BUT BUT BUT BUT BUT BUT BUT BUT BUT!!!!

having said that: If you take out 5ml of a concentrated solution (lets say 1 unit) then the concentration will remain the same.. no? Maybe the question you asked is wrong.. infact the concetration will remain the same no matter how much of the solution you will take out!

I understand what your saying but if you take out say 5ml and 25ml of ater to it, it wll get diluted down so shouldnt the concentration change?????

Nd is Neodyumium, yh f block element.

[Guest _guest_]

Neodyumium

Just never used in solutions.

Is it really able to form solution?

I have not find rare elements used in solutions as said.

Really i am confused

I was thinking Neodyumium is not soluble under normal conditions

Chemistry is too much for me.

Is element genuinely used or just used for a question hypothetical.

I dont get its use insolution:(

[satty2007]

Neodyumium

Just never used in solutions.

Is it really able to form solution?

I have not find rare elements used in solutions as said.

Really i am confused

I was thinking Neodyumium is not soluble under normal conditions

Chemistry is too much for me.

Is element genuinely used or just used for a question hypothetical.

I dont get its use insolution:(

it was actually neodymium oxide, and not neodymium metal

[Guest _Guide_]

Just using Neodyumium as heavy metal as it is present in F block.

I don't think it makes solution as explained. I was thinking it is not soluble in water.

Are you sure it is soluble.

[satty2007]

Just using Neodyumium as heavy metal as it is present in F block.

I don't think it makes solution as explained. I was thinking it is not soluble in water.

Are you sure it is soluble.

Ok, i used Neodymium oxide to make the solutions and it was dissolved in concentrated nitric acid not water.

Hope that clears things up.

[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.