Answer to Question #138908 in General Chemistry for Youssef Ahmed Ismail

Question #138908

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?


1
Expert's answer
2020-10-19T14:21:02-0400

First we need to determine the number of moles of Ba(OH)2 in the initial solution:

"n(Ba(OH)_2)=\\frac{74.28g}{171.34g\/mol}=0.4335mol"

In order to prepare 1.000 L of 0.100M Ba(OH)2 solution, 0.100 moles of Ba(OH)2 are required.

Therefore, the volume of the initial solution that must be diluted equals:

"V=2.450L\\times\\frac{0.100mol}{0.4335mol}=0.565L=565mL"


Answer: 565 mL


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