A 26 mL sample of a solution of MgF2 was diluted with water to 75 mL. A 25 mL sa

Question

A 26 mL sample of a solution of MgF2 was diluted with water to 75 mL. A 25 mL sample of the dilute solution was found to contain 0.04 moles of F-. What was the concentration of MgF2 in the original undiluted solution?

Explanation / Answer

MgF2 =====> Mg2+ + 2 F- So the molar ratio is 1 MgF2 : 2 F-. so 0.04 mol F- * ( 1 mol MgF2 / 2 mol F-) = 0.02 mol MgF2 Now Molarity = moles /volume Molarity of MgF2 = 0.02 mol /(25/1000)L=0.8 M MgF2 The dilution formula is M1V1=M2V2 M1= molarity of undiluted MgF2 = ? : V1 = volume of undiluted MgF2 = 26 mL M2 = molarity of diluted MgF2 = 0.08 M : V2 = diluted volume = 75mL M1 = (M2V2)/V1 = (0.08*75)/26=0.2307 M = 0.231 M

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