[jonj1_12]

I have 1000ml of 0.5M stock solution of sodium chloride. I need to dilute this to make 200ml at a concentration o 0.2M solution. What volumes of sodium chloride do i need to add and what volume of water do i need to add to make up the 200ml?

Thank you if you can help

[Hunter2]

Caculate how much mole you have in stock and also how much is in the new solution. Then calculate the ratio.

The ratio of the molarities  =  ratio of the volumes

[jonj1_12]

Is this correct?
Number of mols = concentration x volume
= 0.5 x 1L = 0.5mol
= 0.2 x 0.2L = 0.04mol

(I change the ml to L as i was always under the impression that you always deal with L..)

0.5/0.04 = 12.5
So is it 12.5ml of sodium chloride and 187.5ml of water?
Is this correct?
Thank you for the *delete me*

[jonj1_12]

Thanks for the help

[Hunter2]

No its wrong. I said the ratios are equal.

The ratio is 1:12.5 is correct.  So you have to calculate the new volume. Start volume is 1 l.

[jonj1_12]

ok, can you show me your workings out with answer so i can see how it is done more clearly please.

thanks

[Hunter2]

If you know the ratio then you can work out the volume. 1:12.5 = x ml /1000 ml, x = 80 ml. This is your NaCl solution.

So how much water you need?

[Borek]

How many moles of NaCl in the target solution? In how what volume of the stock solution do you have this amount of NaCl?

[killazerg]

If this is a lab question, why not use the serial dilution method?

Use this equation to solve the problem: 

M₁V₁ = M₂V₂

We know:

M₁ = 0.5M
V₁ = 1 L
M₂ = 0.2M
V₂ = ?

By solving algebraically, you get V₂ = 2.5 L. You would have to dilute the current solution up to 2.5 L by adding 1.5 L of of water. From there, you can take .200 L of 0.2 M NaCl

[Hunter2]

But nobody will do it in this way. To waste to much chemistry.
In the third post the moles already calculated. The ratio is 1:12.5 what means from the 1 l stock solution 80 ml has to be taken, immersed into a 200 ml flask and filled up with water approx. 120 ml.