A student obtains 100 mL of a 0.0003627 M solution of AgNO_3 (silver nitrate). H
ID: 492073 • Letter: A
Question
A student obtains 100 mL of a 0.0003627 M solution of AgNO_3 (silver nitrate). He labels this "solution #1". He then pipets 5 mL of Solution #1 into a 50 ml volumetric flask and dilutes to the mark with water. He labels this "Solution #2." He then pipets 10 mL of Solution #1 into a 250 mL volumetric flask and dilutes to the mark with water. This is "Solution #3". Finally, the prepares 'Solution #4" by pipetting 16 mL of solution #3 into a 50 mL volumetric flask and dilutes to the mark with water. Draw relational pictures to illustrate the relationship between the original solution and all of the diluted solutions that the student prepared. What is the concentration of silver nitrate in Solution #4? Consider the silver nitrate dilution calculation question you just answered. Match each to the appropriate answer. Which solution is both a stock solution AND a dilution? Which solution is only a stock solution? Which solution is a 1:10 dilution of Solution #1? the concentration of Solution #2 should be reported to this many sig figs: the concentration of Solution #3 should be reported to this many sig figs: the concentration of Solution #4 should be reported to this many sig figs: Which solution is a 1:25 dilution of Solution #1?Explanation / Answer
M1V1 = M2V2
M1 = 0.0003627 M
V1= 5.00 ml
M2 = concentration of solution 2
V2= 50 ml volume of solution 2
M1V1= M2V2
0.0003627*5.00 = M2*50 ML
M2= 3.627*10^-5 M
Now same fro solution 3
M2V2 = M3V3
3.627*10^-5 M* 10 ml = M3*250
M3= 1.4508*10^-6 M
Then solution 4
M3V3= M4V4
1.4508*10^-6 M *16 ml = M4*50 ml
M4 = 4.64*10^-7 M
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