Explanation To calculate the density of the solution Record the given data Volume of H 3 PO 4 = 100 .0 mL Mass of H 3 PO 4 = 10 .0 g Density of water = 1 .00g/cm 3 100 mL of water has a mass of 100 g since the density of water is 1 .00g/cm 3 Density of solution = 10 .0 g H 3 PO 4 +100 .0 g H 2 O 104 mL = 1 .06 g cm -3 Density of solution = 1 .06 g cm -3 To calculate the mole of H 3 PO 4 and Water: Record the given data Mass of H 3 PO 4 = 10 .0 g Mass of Water = 100 .0 g Molar mass of H 3 PO 4 = 97 .99 g Molar mass of Water = 18 .01 g Moles of H 3 PO 4 = 10 .0 g× 1 mol 97 .99 g = 0 .102 mol Moles of Water = 100 .0 g× 1 mol 18 .02 g = 5 .55 mol Moles of H 3 PO 4 = 0

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ISBN: 8220100552236

Author: ZUMDAHL

Publisher: CENGAGE L

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Chapter 10, Problem 29E

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The density, molality, molarity and mole fraction of solution has to be calculated.

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Chapter 10, Problem 28Q

Chapter 10, Problem 30E

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The density, molality, molarity and mole fraction of solution has to be calculated.

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Interpretation: The density, molality, molarity and mole fraction of solution has to be calculated.

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Chapter 10 Solutions

Interpretation:

The density, molality, molarity and mole fraction of solution has to be calculated.