High School

A student wishes to prepare 2.00 liters of a 0.100 M KIO₃ (molecular weight 214). The proper procedure is to weigh out:

A. 42.8 grams of KIO₃ and add 2.00 kg of H₂O
B. 42.8 grams of KIO₃ and add H₂O until the final homogeneous solution has a volume of 2.00 L
C. 21.4 grams of KIO₃ and add H₂O until the final homogeneous solution has a volume of 2.00 L
D. 42.8 grams of KIO₃ and add 2.00 liters of H₂O
E. 21.4 grams of KIO₃ and add 2.00 liters of H₂O

Answer :

Final answer:

To prepare a 2.00 L solution of 0.100 M KIO₃, you need to weigh out, dissolve 42.8 grams of KIO₃ and then add water until the total volume of the solution is 2.00 L. this is because the volume of the solvent can increase after solute is added.

Explanation:

To answer your question, a student wishes to prepare 2.00 Liters of a 0.100 M KIO₃ (molecular weight 214). First, we need to calculate the amount of KIO₃ you need to create this solution. the number of moles needed is determined by the molarity of the solution times the volume, in liters. therefore, moles = Molarity * Volume = 0.100 M * 2.00 L = 0.200 mol KIO₃.

Since the molecular weight of KIO₃ is 214 g/mol, we multiply the moles by the molecular weight to get the mass. thus, Mass = moles * molecular weight = 0.200 mol * 214 g/mol = 42.8 grams KIO₃.

Therefore, the proper procedure would be (b) 42.8 grams of KIO₃ and add H₂O until the final homogeneous solution has a volume of 2.00 L. it's important to add water until the final volume is 2.00 L because adding the solute can increase the volume of the solvent.

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