How many mililiters of 0.1126 M Ba(NO3)2 contain 11.45 g of Ba(NO3)2

Chemistry: Principles and Practice
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Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
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Chapter4: Chemical Reactions In Solution
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How many mililiters of 0.1126 M Ba(NO3)2 contain 11.45 g of Ba(NO3)2

Expert Solution
Step 1

Given : concentration of Ba(NO3)2 = 0.1126 M

And mass of Ba(NO3)2 = 11.45 g

Molar mass of Ba(NO3)2 = Atomic mass of Ba + Atomic mass of N X 2 + Atomic mass of O X 6 

=> Molar mass of Ba(NO3)2 = 137.3 + 14 X 2 + 16 X 6 = 261.3 g/mol

 

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