You need to prepare 500.00 mL of a 0.25 M solution of nitric acid . Concentrated nitric acid comes from the chemical copmanies as a 15.8 M liquid. How much (in mL) of this 15.8 M solution will you need to dilute to 500.00 mL to get the desired 0.25 M solution?
The question as posed makes little sense in terms of significant figures and practicalities. You would not use glassware of 500.00 ml +/- 0.01 ml (5 sig figs) to prepare a solution of 0.25 M (2 sig figs). Incidently a standard 500 ml class A volumetric glassware has an accuracy of +/- 0.15 ml.
I digress...
to get 500 ml of a 0.25 M solution you need 0.125 mole of nitric acid made up to 500 ml total volume.
To find the volume (in ml) of nitric acid needed we can do this by simple proportion.
15.8 mole in 1000 ml
so
1.00 mole in 1000 ml/15.8
and
0.125 mole in 1000 x 0.125 ml/15.8
which you only need to quote to 2 sig figs
To find out how much of the 15.8 M solution you need to dilute, you can use the formula:
(M1)V1 = (M2)V2
Where:
M1 = initial concentration of the solution (15.8 M)
V1 = volume of the initial solution (unknown)
M2 = desired concentration of the final solution (0.25 M)
V2 = final volume of the solution (500.00 mL)
Rearranging the formula, we get:
V1 = (M2 * V2) / M1
Now, let's substitute the given values into the formula:
V1 = (0.25 M * 500.00 mL) / 15.8 M
V1 = 7.91 mL
Therefore, you would need to dilute approximately 7.91 mL of the 15.8 M solution with enough water to make a total volume of 500.00 mL to obtain a 0.25 M solution of nitric acid.