A 74.28-g sample of Ba(OH)₂ is dissolved in enough water to make 2.450 liters of solution.How many mL of this solution must be diluted with water in order to make 1.000 L of 0.100 M Ba(OH)₂?

Question

Quizplus · Accepted Answer

First, determine the molarity of the original solution. The molar mass of Ba(OH)2 is approximately 171.34 g/mol (137.33 for Ba + 2*(16 for O + 1 for H)). Thus, 74.28 g of Ba(OH)2 is 74.28 g / 171.34 g/mol = 0.433 mol. The molarity of the original solution is 0.433 mol / 2.450 L = 0.177 M. Using the dilution formula M1V1 = M2V2, where M1 is the initial molarity (0.177 M), V1 is the volume of the solution needed, M2 is the final molarity (0.100 M), and V2 is the final volume (1.000 L), we can solve for V1: V1 = (M2V2) / M1 = (0.100 M * 1.000 L) / 0.177 M = 0.565 L, which is 565 mL.

A 7428-G Sample of Ba(OH)2 Is Dissolved in Enough Water to to Make

A 7428-G Sample of Ba(OH)₂ Is Dissolved in Enough Water to to Make