how could you prepare 60 cm3 of a 0.20 mol dm-3 solution of nitric acid from stock solution 4.00 mol dm-3 acid

Answer 1

well divide 4.00 by a hundred to get 1% of it.
which is 0.04 mol, so you'd need 5% to get up to 0.20

work out 5% of 60cm3, thats how much of the nitric acid you'd need to add. So that'd be 3 cm3 of nitric acid.
To get the 60cm3 fill it up with water, that'll get the molarity down to correct levels. 57cm3 of water.

water is neutral so it won't effect the experiment unless its not clean water.

Answer 2

same as above:

use the equation C1V1=C2V2

C1 = stock solution conce
V1= volume of stock

C2= Concentration you want
V2= volume want of C2

so pu in the numbers

C1V1=C2V2

4.00 mol dm-3 x V1(the unkown) = 0.20mol dm-3 x 60

rearrange 0.20 mol x 60 / 4.00 = V1 so V1= 3cm3

So you need 3 cm3 of stock solution which you add 57 cm3 to to maske uo to 60.

This sounds long winded to write down but is easy C1V1=C2V2

Answer 3

Add 3 cm3 stock solution to 57 cm3 water.
[1+19] because 4.00*0.20=20=[1+19]
therefore, maintaining ratio;
3 cm3 stock is added to 57 cm3 H2O
to give the required amount
at the correct dilution.