What volume of 0.200 M silver nitrate solution (AgNO3) would have to be diluted to form 500mL of 0.050 M solution?
you are diluting it .2/.05=4 times. You need one part the original stock solution, and three parts water.
so, four parts of 500ml gives each part as 125ml.
thanks very much for your help, have a bless evening
To solve this question, we can use the formula for dilution:
C1V1 = C2V2
Where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume. In this case, we are given C1 = 0.200 M, C2 = 0.050 M, and V2 = 500 mL.
Now, let's calculate V1, the initial volume of the 0.200 M solution that needs to be diluted.
C1V1 = C2V2
(0.200 M)(V1) = (0.050 M)(500 mL)
To calculate V1, rearrange the formula:
V1 = (C2V2) / C1
V1 = (0.050 M)(500 mL) / (0.200 M)
Now, let's do the actual calculation:
V1 = (0.050 M)(500 mL) / (0.200 M)
V1 = 125 mL
Therefore, you would need to dilute 125 mL of the 0.200 M silver nitrate solution to form 500 mL of the 0.050 M solution.