Chapter 13, Problem 13.17QP

(a)

Interpretation Introduction

Interpretation: For the aqueous $\text{1}{\text{.22 M Sugar (C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}\text{)}$ solution molality has to be calculated.

Concept introduction:

Molality: Molality is defined as number of moles of the solute present in the specified amount of the solvent in kilograms.

$\text{Molality =}\frac{\text{number of moles of the solute}}{\text{kg of solvent}}$

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $\text{12}\text{g}$ of ${}^{\text{12}}\text{C}$.

From given mass of substance moles could be calculated by using the following formula,

$\text{Moles}\text{of}\text{substance }\text{= }\frac{\text{Given}\text{mass}\text{of}\text{substance}}{\text{Molecular}\text{mass}}$

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$.

Answer to Problem 13.17QP

Molality of $\text{1}{\text{.22 M Sugar (C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}\text{)}$ solution is $\text{1}\text{.74m}$.

Explanation of Solution

Given data: Strength of sugar solution $\text{=1}\text{.22 M}$

Density of sugar solution $\text{=1}\text{.12g /mL}$

Calculation of mass of sugar:

Substitute the value of strength of sugar and molecular mass of sugar in the formula to calculate mass of sugar.

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

$\begin{array}{c}\text{mass of sugar = 1}\text{.22 }\cancel{\text{mol}}\text{ ×}\frac{\text{342}\text{.3g}}{\text{1}\cancel{\text{mol}}\text{ sugar}}\text{= 418 g}\\ \text{418g = 0}\text{.418}\text{kg}\left[\because \text{1kg =1000g}\right]\end{array}$

Calculation of mass of sugar solution:

$\begin{array}{c}\text{mass of solution =1000}\cancel{\text{mL}}\text{×}\frac{\text{1}\text{.12g}}{\text{1}\cancel{\text{mL}}}=\text{1120 g}\\ \text{1120 g = 1}\text{.120}\text{kg}\left[\because \text{1kg =1000g}\right]\end{array}$

Calculation of molality of the solution:

Substitute the value of moles of sugar and amount of solvent (water) into molality formula to calculate molality of sugar solution.

$\text{molality =}\frac{\text{1}\text{.22 mol sugar}}{\text{(1}\text{.120-0}{\text{.418)kg H}}_{\text{2}}\text{O}}\text{= 1}\text{.74m}$

(b)

Interpretation Introduction

Interpretation: For the aqueous $\text{0}\text{.87 M NaOH }$ solution molality has to be calculated.

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.

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

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $\text{12}\text{g}$ of ${}^{\text{12}}\text{C}$.

From given mass of substance moles could be calculated by using the following formula,

$\text{Moles}\text{of}\text{substance }\text{= }\frac{\text{Given}\text{mass}\text{of}\text{substance}}{\text{Molecular}\text{mass}}$

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$.

Answer to Problem 13.17QP

Molality of $\text{0}\text{.87 M NaOH }$ solution is $\text{0}\text{.87 m}$.

Explanation of Solution

Given data: Strength of Sodium hydroxide solution $\text{= 0}\text{.87M}$

Density of Sodium hydroxide solution $\text{=1}\text{.04 g /mL}$

Calculation of mass of$\text{NaOH}$:

Substitute the value of strength of $\text{NaOH}$ and molecular mass of $\text{NaOH}$ in the formula to calculate mass of $\text{NaOH}$.

Molecular weight of $\text{NaOH}$$\text{= 40g}$

$\text{mass of NaOH = 0}\text{.87 }\cancel{\text{mol}}\text{NaOH ×}\frac{\text{40g NaOH}}{\text{1}\cancel{\text{mol}}\text{ NaOH}}\text{= 35g NaOH}$

Calculation of mass of ${\text{H}}_{\text{2}}\text{O}$solution:

$\begin{array}{l}\text{mass of solvent =1040g - 35g =1005g =1}\text{.005kg}\\ \because \text{1kg =1000g}\end{array}$

Calculation of molality of the solution:

Substitute the value of moles of Sodium hydroxide and amount of solvent (water) into molality formula, to calculate molality.

$\text{molality =}\frac{\text{0}\text{.87 mol }}{\text{1}{\text{.005kg H}}_{\text{2}}\text{O}}\text{= 0}\text{.87m}$

(c)

Interpretation Introduction

Interpretation: For the aqueous $\text{5}{\text{.24 M NaHCO}}_{\text{3}}$ solution molality has to be calculated.

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.

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

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $\text{12}\text{g}$ of ${}^{\text{12}}\text{C}$.

From given mass of substance moles could be calculated by using the following formula,

$\text{Moles}\text{of}\text{substance }\text{= }\frac{\text{Given}\text{mass}\text{of}\text{substance}}{\text{Molecular}\text{mass}}$

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$.

Answer to Problem 13.17QP

Molality of $\text{5}{\text{.24 M NaHCO}}_{\text{3}}$ solution is $\text{6}\text{.99m}$.

Explanation of Solution

Given data: Strength of Sodium bicarbonate solution $\text{=5}\text{.24 M}$

Density of Sodium bicarbonate solution $\text{=1}\text{.19g /mL}$

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

Substitute the value of molecular mass and strength of Sodium bicarbonate to calculate mass of ${\text{NaHCO}}_{\text{3}}$.

${\text{mass of NaHCO}}_{\text{3}}\text{ = 5}\text{.24 }\cancel{\text{mol}}\text{ ×}\frac{\text{84}\text{.01g}}{\text{1}\cancel{\text{mol}}{\text{ NaHCO}}_{\text{3}}}{\text{= 440g NaHCO}}_{\text{3}}$

$\begin{array}{l}{\text{mass of solvent (H}}_{\text{2}}\text{O) =1190g - 440g = 750g = 0}\text{.750kg}\\ \because \text{1kg =1000g}\end{array}$

Calculation of molality for given solution:

Substitute the value of moles of Sodium bicarbonate and amount of solvent (water) into molality formula to calculate molality of the given solution.

$\text{molality =}\frac{\text{5}{\text{.24 mol NaHCO}}_{\text{3}}}{\text{0}{\text{.750 kg H}}_{\text{2}}\text{O}}\text{= 6}\text{.99 m}$