What volume would be needed to prepare 375 mL of a .45 M CaCl2 using only a solution of 1.0 M CaCl2 and water?

asked Aug 4, 2022 55.2k views

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7 votes

Answer:

168.75 ml

Step-by-step explanation:

M1V1=M2V2

375ml*.45M=1M*V2

Sator answered Aug 10, 2022

by Sator

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