How many milliliters of 0.580 m aluminum bromide solution are required to prepare 89.3 ml of a solution that is 0.233 m in bromide ion?

The initial volume is 35, 9 ml

Explanation:

We use the formula:

V initial  x C Initial  = V Final  x C Final   , being V = volume and C = concentration

V initial x 0, 580 M = 89, 3 ml x 0, 233 M

V initial = (89, 3 ml x 0, 233 M)/0, 580 M = 35, 9 ml

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