Problem:  Describe how you would prepare 350.0 mL of 0.100 M C12H22O11 starting with 3.00 L of 1.50 M C12H22O11.

FREE Expert Solution

M1V1=M2V2

M1 = initial concentration
V1 = initial volume
M2 = final concentration
V2 = final volume


M1 = 1.50 M C12H22O11
V1 = ?
M2 = 0.100 M C12H22O11
V2 = 350.0 mL

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Problem Details

 Describe how you would prepare 350.0 mL of 0.100 M C12H22O11 starting with 3.00 L of 1.50 M C12H22O11.

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