Diluting a solution, get the right molarity.?

Question

"In car batteries, 6.0 M sulfuric acid is used. If you purchase 18.0 M sulfuric acid, how could you dilute the concentrated acid to get the right molarity if you needed 2.0 liter of the 6.0 M battery acid?"

The calculation:
(2 liters) x (6.0M / 18.0M) = 0.67 L of 6.0M sulfuric acid.

So,
You add 0.67 L of 18.0M sulfuric acid into a flask, add water until the solution is 2L.


Okay, my problem is that I don't understand the calculation. Where does the "(6.0 M / 18.0 M)" come from / what is the reasoning behind everything?

Thanks in advance!

Answer 1

You can use the formula :
M1V1 = M2V2
18.0 x V2 = 6.0 x 2.0
V2 = 6.0 x 2.0 / 18.0

Answer 2

18moles per 1000cc solution
how much solution to have only 6 moles ?
or start with 18m/1000cc
multiply both top and bottom by same number so top is equal to 6
use value of 6/18
take 6/18 times 18 to get 6 then multiply bottom by same number or (6/18) 1000 = 333cc
333cc of 18m acid is 6 moles that added to water to bring volume upto 1000cc is 1 liter 6 moler
for 2 liters use twice as much ie 333 times 2 = 666
666ccs 18moler added water to make 2000cc

Answer 3

Your calculation is wrong---you dilute 0.66666666=0.67L of 18 M Sulphuric acid to 2.0 L of 6M Sulphuric acid(1>>>3 dilution)--M1V1=M2V2

Answer 4

basically use the formula c1v1 = c2v2 (c being your concentration or molarity and v being your quantity.) substitute 4.5 and six.a million for c1 and v1 and substitute fifty 5 liters for v2. Then use algebra to sparkling up.

Answer 5

the formula above that Dr.A has said,is true but it is come from this fact that if u diliute a solotion the number of mols does not change,so we have: no. of mols (of 18M acid)=no. of mols (of 6M acid) ,and C=m/V ,n=m/w(molar weight),so we have: n1V1=n2V2 .