# Simple molar ration problem

1. Dec 29, 2015

### bilalarif

I have been given NiAc and ZnAc solutions of 1M concentration. And being asked to mix the two solutions together in Ni/Zn nominal molar proportions 1%, 3% and 5%. How can I calculate and mix the two solutions to get the required molar proportion.

2. Dec 29, 2015

### Bystander

Ni/Zn = 0.01, 0.03, and 0.05. Calculate it and do it.

3. Dec 30, 2015

### bilalarif

Molar fraction = (number of moles of component of interest) ÷ (total number of moles of mixture)

For Ni/Zn nominal proportion 1%
0.01= number of moles of Ni/ total number of moles of Ni+Zn
Suppose we have total moles of (Ni+Zn) solution= 5 mmol
0.01= moles of Ni/ 5mmol
Moles of Ni = 0.05 mmol
Moles of Zn= total moles of (Ni+Zn)- moles of Ni
Moles of Zn= 4.95 moles
As both NiAc and ZnAc concentration is 1M.
M=n/V
V (ZnAc solution) = 4.95mmole/1 mol/l = 4.95ml

V (NiAc solution) = 0.05mmole/1 mol/l = 0.05ml

is it correct calculation?

Last edited: Dec 30, 2015
4. Dec 30, 2015

### Bystander

That's one. Two to go.

5. Dec 30, 2015

### epenguin

No it isn't, though the error is small. You have made a solution which is 0.01 M Ni, but that is not what was asked for. It is not quite 1M Zn any more, so 1 % is not exactly the molar proportion.
If the starting molarities are equal an you are asked for a 1 % ratio you just mix them in the ratio 100:1.
You're probably so used to problems of achieving a given final molarity that you have overthught this one

6. Dec 30, 2015

### Bystander

Divalent ions, one to one.