Math question

[Hadi]

Hi friends,

I am little bit confused on a math question :

 

How many mL of water should be added to 200 ml of 1:125 w/v solution to make a solution such that 50 ml diluted to 100 ml will provide 1:4000 dilution?

 

The answer was 3000 mL

 

Any help appreciated....

[anilchava]

U need to prepare final solution of 1:4000 by diluting 50ml to 100mls.so here dilution factor is 2x.So very simply you should require a 1:2000 dilution from 1:125 stock solution.

200/125 =x/2000

x=3200ml

Water to be added =3200-200

=3000ml

[Hadi]

Thanks for the reply...

 

But what you do if the dilution factor is not 2x...that what confuses me

 

see this slightly different example,

How many mL of water should be added to 100 ml of 1:10 w/v solution to make a solution such that 20 ml diluted to 100 will provide a 1:2000 dilution?

 

For the first example i posted...i thought about it in a different way:

 

For original solution of strength 1:125 the percentage strength is 0.8%

and the final solution of strength 1:4000 the percentage strength is 0.025%

Now C1V1=C2V2

(0.8)(100)=(0.025)V2

V2=3200 ml

Amount to be added to water= 3200-200=3000 ml

 

But when i try to answer the other example similarly i got different answer

i got 19900 ml to be added while it is supposed to be 3900 ml !

[Lenny32]

Hello Hadi,

With regards to your question,always work from the back to the front.

First of all,you get the final solution conc.1:2000 = 0.05%w/v then find out the conc. of the 20ml that would be diluted to 100ml:

use C1V1 = C2V2

C1=?, C2= 0.05%, V1=20ml, V2= 100ml thus C1 = 0.25w/v

After that find the conc. of 1:10 solution = 10%w/v

2nd stage

and then C1 = 10%w/v. V1=100ml, C2 would now be =0.25%, V2=?

 

10% x 100ml = V2 V2 = 4000ml thus water added = 4000ml - 100ml =

0.25% = 3900mls

 

I hope that this helps

[resu]

hi lenny32,

 

i am also confused here. yes, you are right,but if the formulla for question 1 worked then it should be worked on question 2 too.

 

 

as, here question style is same for both , then formulla should be same. otherwise, in exam how one can predict that we have to use different formulla for each question?

 

also tell me how one can find this dilution factor?

please guide me.:rolleyes: :mad:

[anilchava]

In your second probelm dliution factor is 5x. so you should require a 1:400 dilution from 1:10 stock solution.

100/10 =x/400

=4000ml

Water to be added =4000-100

=3900ml

[Hadi]

Thank you guys...that was really helpful.

[newfpgee]

hi anil how did u get 100 in the equation 100/10 =x/400

how did u put dilution factor into the equation.

and cud u tell me which question in manan calculations is this question?

[Cuteiii]

How about this

How many milliliters of water should be added to 100 ml of a 1:125 w/v solution to make a solution such that 25 mL diluted to 100 mL will yield a 1:4000 dilution?