Chapter 13, Problem 13.19QP

Calculate the molalities of the following aqueous solutions: (a) 0.988 M sugar (C₁₂H₂₂O₁₁) solution (density of solution = 1.095 g/mL), (b) 1.12 M NaOH solution (density of solution = 1.082 g/mL), (c) 2.98 M NaHCO₃ solution (density of solution = 1.077 g/mL).

Interpretation Introduction

Interpretation: For the following aqueous solutions molality to be calculated.

$\text{0}{\text{.988 M Sugar (C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}\text{)}$ solution

Concept introduction:

Molality (m): Molality is the number of moles of solute present in one kilogram of solvent. Molality is estimation of moles in relationship with solvent in the solution.

$\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}$

Answer to Problem 13.19QP

Molality of $\text{0}{\text{.988 M Sugar (C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}\text{)}$ solution = $\text{1}\text{.31 m}$

Explanation of Solution

Given data

Strength of sugar solution = $\text{0}\text{.988 M}$

Density of sugar solution = $\text{1}\text{.095g /mL}$

Molality of the solution =?

Calculation of mass of sugar

Molecular weight of sugar = $\text{342}\text{.3g/mol}$

$\text{mass of sugar = 0}\text{.988 mol ×}\frac{\text{342}\text{.3g}}{\text{1mol sugar}}\text{= 338g}$

$\text{338g ×}\frac{\text{1kg}}{\text{1000g}}\text{= 0}\text{.338kg}$

By plugging in the value of strength of sugar and molecular mass of sugar, mass of sugar has calculated.

Calculation of mass of sugar solution

Each case consider 1 L of solution

$\text{mass of 1L solution=1000mL×}\frac{\text{1}\text{.09g}}{\text{1mL}}=\text{1095g}$

$\text{1095g ×}\frac{\text{1kg}}{\text{1000g}}\text{=1}\text{.095kg}$

By multiplying the value of density of sugar and one litter of sugar solution, mass of one liter solution has been calculated.

Calculation of molality of the solution

$\text{molality =}\frac{\text{0}\text{.988 mol sugar}}{\text{(1}\text{.095-0}{\text{.388)kg H}}_{\text{2}}\text{O}}\text{= 1}\text{.31m}$

By plugging the value of moles of sugar and amount of solvent (water) into molality formula, the molality of the solution has calculated.

Interpretation Introduction

Interpretation: For the following aqueous solutions molality to be calculated.

$\text{1}\text{.12 M NaOH }$ Solution

Concept introduction:

Molality (m): Molality is the number of moles of solute present in one kilogram of solvent. Molality is estimation of moles in relationship with solvent in the solution.

$\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}$

Answer to Problem 13.19QP

Molality of $\text{1}\text{.12 M NaOH }$ solution = $\text{1}\text{.08 m}$

Explanation of Solution

Given data

Strength of Sodium hydroxide solution = $\text{1}\text{.12 M}$

Density of Sodium hydroxide solution = $\text{1}\text{.082g /mL}$

Molality of the solution =?

Calculation of mass of $\text{NaOH}$ and solvent

$\text{mass of NaOH = 1}\text{.12 mol ×}\frac{\text{40g}}{\text{1mol NaOH}}\text{= 44}\text{.8g}$

$\text{mass of solvent =1082g - 44}\text{.8g =1037g =1}\text{.037kg}$

By plugging the value of molecular mass and strength of Sodium hydroxide, mass of $\text{NaOH}$ has calculated.

Calculation of molality of the solution

$\text{molality =}\frac{\text{1}\text{.12 mol }}{\text{1}{\text{.037 kg H}}_{\text{2}}\text{O}}\text{= 1}\text{.08m}$

By plugging the value of moles of Sodium hydroxide and amount of solvent (water) into molality formula, the molality of the solution has calculated.

Interpretation Introduction

Interpretation: For the following aqueous solutions molality to be calculated.

$\text{2}{\text{.98 M NaHCO}}_{\text{3}}$ Solution

Concept introduction:

Molality (m): Molality is the number of moles of solute present in one kilogram of solvent. Molality is estimation of moles in relationship with solvent in the solution.

$\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}$

Answer to Problem 13.19QP

Molality of $\text{2}{\text{.98 M NaHCO}}_{\text{3}}$ solution = $\text{3}\text{.60 m}$

Explanation of Solution

Given data

Given data

Strength of Sodium bicarbonate solution = $\text{2}\text{.98 M}$

Density of Sodium bicarbonate solution = $\text{1}\text{.077g /mL}$

Molality of the solution =?

Calculation of mass of ${\text{NaHCO}}_{\text{3}}$ and solvent

${\text{mass of NaHCO}}_{\text{3}}\text{ = 2}\text{.98 mol ×}\frac{\text{84}\text{.01g}}{{\text{1mol NaHCO}}_{\text{3}}}\text{= 250g}$

${\text{mass of solvent (H}}_{\text{2}}\text{O) =1077g - 250g = 827g = 0}\text{.827kg}$

By plugging the value of molecular mass and strength of Sodium bicarbonate, mass of ${\text{NaHCO}}_{\text{3}}$ has calculated.

Calculation of molality of the solution

$\text{molality =}\frac{\text{2}\text{.98 mol }}{\text{0}{\text{.827 kg H}}_{\text{2}}\text{O}}\text{= 3}\text{.60 m}$

By plugging the value of moles of Sodium bicarbonate and amount of solvent (water) into molality formula, the molality of the solution has calculated.