Chapter U4.83, Problem 7E

Interpretation Introduction

Interpretation:

The solution that would weigh the most needs to be identified from 1.0 L of 1.0 M NaBr, 500mL of 1.0 M KCl and 1.0 L of 0.5 M NaOH.

Concept Introduction:

Molarity is defined as the moles of solute in a liter volume of a solution.

  $\begin{array}{l}\text{Molarity = }\frac{\text{moles of solute}}{\text{L of solution}}\text{ -----(1)}\\ \text{moles = }\frac{\text{mass of solute (m)}}{\text{molar mass of solute (mol}\text{. wt)}}\text{ -----(2)}\\ \text{i}\text{.e}\text{.}\\ \text{Molarity = }\frac{\text{mass of solute}}{\text{molar mass }\times \text{L of solution}}\\ \text{i}\text{.e}\text{. M = }\frac{\text{m}}{\text{mol}\text{.wt × V}}\text{ ------(3)}\end{array}$

The total mass of solution is the sum of the mass of solute and that of the solvent which is water for aqueous solutions

  $\text{Mass of solution = Mass of solute + Mass of solvent (water) ----(4)}$

Answer to Problem 7E

1.0 L of 1.0 M NaBr would weigh the most

Explanation of Solution

Given:

Volume (V) of NaBr = 1.0 L

Molarity (M) of NaBr = 1.0M

Calculation:

Based on equation (1) and (2):

  $\text{Molarity = }\frac{\text{mass of solute}}{\text{molar mass }\times \text{L of solution}}$

Molar mass of NaBr = 102.89 g/mol

  $\begin{array}{l}\text{Mass NaBr = 1}\text{.0 mol/L}\times \text{102}\text{.89 g/mol }\times \text{1}\text{.0 L}\\ \text{= 103 g}\end{array}$

Based on equation (4):

  $\begin{array}{l}\text{Mass of solution = Mass of solute + Mass of solvent (water) }\\ \text{Density of water = 1g/ml}\\ \text{For a volume = 1}\text{.0 L (1000 ml), the mass of water = 1000 g}\\ \text{Mass of solution = 103 + 1000 = 1103 g}\end{array}$

b)

Given:

Volume of KCl= 500 ml

Molarity of KCl= 1.0 M

Calculation:

Based on equation (1) and (2):

  $\text{Molarity = }\frac{\text{mass of solute}}{\text{molar mass }\times \text{L of solution}}$

Molar mass of KCl = 74.55 g/mol

  $\begin{array}{l}\text{Mass KCl = 1}\text{.0 mol/L}\times \text{74}\text{.55 g/mol }\times \text{0}\text{.500 L}\\ \text{= 37}\text{.3 g}\end{array}$

Based on equation (4):

  $\begin{array}{l}\text{Mass of solution = Mass of solute + Mass of solvent (water) }\\ \text{Density of water = 1g/ml}\\ \text{For a volume = 500 ml, the mass of water = 500 g}\\ \text{Mass of solution = 37}\text{.3 + 500 = 537 g}\end{array}$

c)

Given:

Volume of NaOH = 1.0 L

Molarity of NaOH = 0.5 M

Calculation:

Based on equation (1) and (2):

  $\text{Molarity = }\frac{\text{mass of solute}}{\text{molar mass }\times \text{L of solution}}$

Molar mass of NaOH = 40 g/mol

  $\begin{array}{l}\text{Mass NaOH = 0}\text{.5 mol/L}\times \text{40 g/mol }\times \text{1}\text{.0 L}\\ \text{= 20 g}\end{array}$

Based on equation (4):

  $\begin{array}{l}\text{Mass of solution = Mass of solute + Mass of solvent (water) }\\ \text{Density of water = 1g/ml}\\ \text{For a volume = 1000 ml, the mass of water = 1000g}\\ \text{Mass of solution = 20 + 1000 = 1020 g}\end{array}$

Conclusion

Therefore, based on the calculations, 1.0 L of 1.0 M NaBr would weigh the most