1) In the laboratory, a student dilutes 18.3 mL of a 9.98 M perchloric acid solu
ID: 815663 • Letter: 1
Question
1) In the laboratory, a student dilutes 18.3 mL of a 9.98 M perchloric acid solution to a total volume of 300.0 mL. What is the concentration of the diluted solution ?
------> ANSWER: _________ M
2) How many milliliters of 11.1 M hydroiodic acid solution should be used to prepare 3.50 L of 0.600 M HI?
------> ANSWER: _________ mL
3) In the laboratory you dilute 2.47 mL of a concentrated 12.0 M nitric acid solution to a total volume of 150 mL. What is the concentration of the dilute solution?
-------> ANSWER: __________ M
4) You wish to make a 0.405 M perchloric acid solution from a stock solution of6.00 M perchloric acid. How much concentrated acid must you add to obtain a total volume of 100 mL of the dilute solution?
--------> ANSWER: _________ mL
Explanation / Answer
V1= 18.3
1)
M1= 9.98
V2 = 300
M2 =
Use this formula
M1*V1 = M2*V2 (this comes from am ole balance, molarity times volume gives you the number of moles, which dont change when diluting
18.3ml*9.98 M = 300 ml *M
M = 0.6 M
2)
Similar to past question...
M1*V1 = M2*V2
11.1 M * V1 = 0.6 M * 3.5 L
V1 = 0.6*3.5 / 11.1 = 0.18 Liters or 180 ml
3) Similar to past questions
M1*V1 = M2 * V2
2.47 ml * 12M = 150 ml * M2
M2 = 2.47/150 * 12 = 0.197 M
4) Once again, the same questions
M1*V1 = M2*V2
0.405 M * V1 = 6M*100 ml
V1 = 6/0.405*100 = 1481 ml or 1.4 liters
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