The following questions all refer to an aqueous 4.12 M iron(II) nitrate stock so
ID: 1056045 • Letter: T
Question
The following questions all refer to an aqueous 4.12 M iron(II) nitrate stock solution. If 24.5 mL of the stock solution are diluted to 786 mL, what is the molarity (in mol/L) of the new solution? If 20.7 mL of the stock solution are diluted to obtain a 0.627 M iron(II) nitrate solution, what is the volume (in mL) of the new solution? Determine the volume (in mL) of the original stock solution that must be used to prepare 842 mL of a 0.233 M iron(II) nitrate solution. Determine the volume of water (in mL) that was added to 28.2 mL of the stock solution to prepare a 0.191 M iron(II) nitrate solution.Explanation / Answer
we will use dilution formula:
M1*V1 = M2*V2
1---> is for stock solution
2---> is for diluted solution
M1*V1 = M2*V2
1)
M1*V1 = M2*V2
4.12*24.5 = M2*786
M2 = 0.128 M
2)
M1*V1 = M2*V2
4.12*20.7 = 0.627*V2
V2 = 136 mL
3)
M1*V1 = M2*V2
4.12*V1 = 0.233*842
V1 = 47.6 mL
4)
M1*V1 = M2*V2
4.12*28.2 = 0.191*V2
V2 = 608 mL
volume of water added= 608 mL - 28.2 mL = 579.8 mL
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