How many milliliters of a stock solution of 8.50 M MgCl2 would you have to use to prepare 650.0mL of a 1.60 M MgCl2 solution?
We will use the formula
M1V1 = M2V2
8.5 x V1 = 1.60 x 0.650
V1 = 0.122 L or 122.35 mL of stock solution we would require in this case
How many milliliters of a stock solution of 8.50 M MgCl2 would you have to use...
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