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mojhsa [17]
3 years ago
10

15. What volume of water must be added to 300 mL of 0.75 M HCl to dilute the solution to 0.25 M?

Chemistry
1 answer:
Sliva [168]3 years ago
7 0

Known :

V1 = 300 mL

M1 = 0.75 M

M2 = 0.25 M

Solution :

M1 • V1 = M2 • V2

(0.75 M) • (300 mL) = (0.25 M) V2

V2 = 900 mL

Water add to this solution is :

∆V = V2 - V1

∆V = 900 - 300

∆V = 600 mL

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Answer:

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You need to make an aqueous solution of 0.182 M aluminum sulfate for an experiment in lab, using a 250 mL volumetric flask. How
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Answer:

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