chemistry problem?

Question

How would you prepare 60 mL of a 0.200M HNO3 from a stock solution of 4.00M HNO3?

Answer 1

You need to equalize the moles of HNO3 for each solution. To do this, use the following equation:
M1*V1 = M2*V2 where M1 and V1 are the molarity and volume of the fist solution and M2 and V2 are the molarity and volume of the second solution.

In this case M1 = 0.200 mol/l, V1 = 0.060, and M2 = 4.00 mol/l

0.200 mol/l *0.060 l = 4.00 mol/l * V2
V2 = 0.003 l
so you would take 57 ml of H2O and to this add 3 mls of 4 M HNO3 to get 60 ml of a 0.2 M solution

Answer 2

M1V1 = M2V2

m1 = 0.200M HNO3
V1 = 60 mL
M2 = 4.00M

Plug and chug.

Answer 3

x(4.00 g/L) = (60 ml)(0.200 g/L)
x = 3 ml
Use 57 ml H20 and 3ml of stock solution.