Question

Question 30 of 38 > A 62.0 mL aliquot of a 1.20 M solution is diluted to a total volume of 238 ml. A 119 ml. portion of that solution is diluted by adding 187 mL of water. What is the final concentration? Assume the volumes are additive. concentration:

Answer 1

Molarity = Number of moles Volume in Liters so number of moles (n) = Molarity (M) ~ volume (V) in litres For dilution of a solution we use the formula Mix V1 = M2 V2 Because the number of moles is constant before dilution and after dilution. Mi = initial Molarity = 1.2 M Vi = initial volume = 62.0 mL M2 = diluted Molarity = M2 M V2 = diluted volume = 238 ml Mi * V1 = M2 * V2 1.2 x 62 = M2 x 238 M2 = 0.313 M2 = diluted Molarity = 0.313 M = 0.313 M = 119.0 mL = M2 M 187 + 119.0= 306.0 mL again diluted M2 = diluted Molarity V1 = diluted volume M3 = final Molarity V3 = final volume M2 * V1 = M3 x V3 0.31 x 119 = M2 306 M2 = 0.122 M2 = final Molarity M2 = final concentration = = 0.122 M 0.122 M.