Chapter 13, Problem 57GQ

Dimethylglyoxime [DMG, (CH₃CNOH)₂] is used as a reagent to precipitate nickel ion. Assume that 53.0 g of DMC has been dissolved in 525 g of ethanol (C₂H₅OH).

s

The red, insoluble compound formed between nickel(II)ion and dimethylglyoxime (DMG) is precipitated when DMG is added to a bask solution of Ni²⁺ (aq).

1. (a) What is the mole fraction of DMG?
2. (b) What is the molality of the solution?
3. (c) What is the vapor pressure of the ethanol over the solution at ethanol’s normal boiling point of 78.4 °C?
4. (d) What is the boiling point of the solution? (DMG does not produce ions in solution.) (K_bp for ethanol = +1.22 °C/m)

Interpretation Introduction

Interpretation: Mole fraction of DMG has to be calculated.

Concept introduction:

Raoult’s law: In a solution, vapor pressure of solvent is proportional to its mole fraction.

  $P_{solvent}=X_{solvent}{P}^0{}_{solvent}$

  where,

  $P^0{}_{solvent}$ is the vapor pressure of pure solvent.

Mole fraction: Amount of that component divided by the total amount of all of the components of the mixture $\left({\text{ n}}_{\text{A}}{\text{ + n}}_{\text{B}}{\text{ + n}}_{\text{C }}\text{+ }\mathrm{...}\right)$.

  ${\text{Mole fraction of A (χ}}_{\text{A}}\text{)= }\frac{{\text{n}}_{\text{A}}}{{\text{ n}}_{\text{A}}{\text{ + n}}_{\text{B}}{\text{ + n}}_{\text{C }}\text{+ }\mathrm{...}}$

Answer to Problem 57GQ

Mole fraction of DMG is $0.0385$

Explanation of Solution

Given,

$\begin{array}{c}\text{Mass}\text{of}\text{ethanol}\text{=}\text{53}\text{.0}\text{g}\\ \text{Mass}\text{of}\text{DMG}\text{=}\text{525}\text{g}\\ \text{Molar}\text{mass}\text{of}\text{ethanol}\text{=}\text{46}\text{.07}\text{g/mol}\\ \text{Molar}\text{mass}\text{of}\text{DMG}\text{=}\text{116}\text{.12}\text{g/mol}\end{array}$

The number of moles of any substance can be determined using the equation

  $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

Number of moles of dimethylglyoxime is,

  $\begin{array}{c}{\text{n}}_{\text{DMG}}\text{=}\frac{\text{53}\text{.0}\text{g}}{\text{116}\text{.12}\text{g/mol}}\\ =0.4564mol\end{array}$

Number of moles of ethanol is,

  $\begin{array}{c}{\text{n}}_{ethanol}\text{=}\frac{\text{525}\text{g}}{\text{46}\text{.07}\text{g/mol}}\\ =11.4mol\end{array}$

Mole fraction of DMG is calculated

$\begin{array}{c}Molefraction,X_{DMG}=\frac{{\text{n}}_{\text{DMG}}}{{\text{n}}_{\text{DMG}}+{\text{n}}_{\text{ethanol}}}\\ =\frac{0.4564mol}{\left(0.4564mol\right)+\left(11.4mol\right)}\\ =0.0385\end{array}$

Mole fraction of DMG is $0.0385$

Interpretation Introduction

Interpretation: Molality of the solution has to be determined.

Concept introduction:

Raoult’s law: In a solution, vapor pressure of solvent is proportional to its mole fraction.

  $P_{solvent}=X_{solvent}{P}^0{}_{solvent}$

  where,

  $P^0{}_{solvent}$ is the vapor pressure of pure solvent.

Mole fraction: Amount of that component divided by the total amount of all of the components of the mixture $\left({\text{ n}}_{\text{A}}{\text{ + n}}_{\text{B}}{\text{ + n}}_{\text{C }}\text{+ }\mathrm{...}\right)$.

  ${\text{Mole fraction of A (χ}}_{\text{A}}\text{)= }\frac{{\text{n}}_{\text{A}}}{{\text{ n}}_{\text{A}}{\text{ + n}}_{\text{B}}{\text{ + n}}_{\text{C }}\text{+ }\mathrm{...}}$

Answer to Problem 57GQ

Molality of the solution is $\text{0}\text{.869}\text{m}$

Explanation of Solution

Given,

$\begin{array}{c}\text{Mass}\text{of}\text{ethanol}\text{=}\text{53}\text{.0}\text{g}\\ \text{Mass}\text{of}\text{DMG}\text{=}\text{525}\text{g}\\ \text{Molar}\text{mass}\text{of}\text{ethanol}\text{=}\text{46}\text{.07}\text{g/mol}\\ \text{Molar}\text{mass}\text{of}\text{DMG}\text{=}\text{116}\text{.12}\text{g/mol}\end{array}$

The number of moles of any substance can be determined using the equation

  $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

Number of moles of dimethylglyoxime is,

  $\begin{array}{c}{\text{n}}_{\text{DMG}}\text{=}\frac{\text{53}\text{.0}\text{g}}{\text{116}\text{.12}\text{g/mol}}\\ =0.4564mol\end{array}$

Molality of the solution is calculated

  $\begin{array}{c}\text{Molality (m) =}\frac{\text{0}\text{.4564}\text{mol}}{\text{0}\text{.525}\text{kg}}\\ \text{=}\text{0}\text{.869}\text{m}\end{array}$

Molality of the solution is $\text{0}\text{.869}\text{m}$

Interpretation Introduction

Interpretation: Mole fraction of DMG and molality of the solution are to be determined. Vapour pressure of ethanol over the solution and Boiling point of the solution are to be calculated.

Concept introduction:

Raoult’s law: In a solution, vapor pressure of solvent is proportional to its mole fraction.

  $P_{solvent}=X_{solvent}{P}^0{}_{solvent}$

  where,

  $P^0{}_{solvent}$ is the vapor pressure of pure solvent.

Mole fraction: Amount of that component divided by the total amount of all of the components of the mixture $\left({\text{ n}}_{\text{A}}{\text{ + n}}_{\text{B}}{\text{ + n}}_{\text{C }}\text{+ }\mathrm{...}\right)$.

  ${\text{Mole fraction of A (χ}}_{\text{A}}\text{)= }\frac{{\text{n}}_{\text{A}}}{{\text{ n}}_{\text{A}}{\text{ + n}}_{\text{B}}{\text{ + n}}_{\text{C }}\text{+ }\mathrm{...}}$

Answer to Problem 57GQ

Vapor pressure of ethanol over the solution is $730.7mmHg$

Explanation of Solution

(c)

Given,

$\begin{array}{c}\text{Mass}\text{of}\text{ethanol}\text{=}\text{53}\text{.0}\text{g}\\ \text{Mass}\text{of}\text{DMG}\text{=}\text{525}\text{g}\\ \text{Molar}\text{mass}\text{of}\text{ethanol}\text{=}\text{46}\text{.07}\text{g/mol}\\ \text{Molar}\text{mass}\text{of}\text{DMG}\text{=}\text{116}\text{.12}\text{g/mol}\\ \text{Vapour}\text{pressure}\text{of}\text{pure}\text{ethanol}\text{at}\text{78}{\text{.4}}^{0}C=760mmHg\end{array}$

The number of moles of any substance can be determined using the equation

  $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

Number of moles of dimethylglyoxime is,

  $\begin{array}{c}{\text{n}}_{\text{DMG}}\text{=}\frac{\text{53}\text{.0}\text{g}}{\text{116}\text{.12}\text{g/mol}}\\ =0.4564mol\end{array}$

Number of moles of ethanol is,

  $\begin{array}{c}{\text{n}}_{ethanol}\text{=}\frac{\text{525}\text{g}}{\text{46}\text{.07}\text{g/mol}}\\ =11.4mol\end{array}$

$\begin{array}{c}Molefraction,X_{ethanol}=\frac{{\text{n}}_{\text{ethanol}}}{{\text{n}}_{\text{DMG}}+{\text{n}}_{\text{ethanol}}}\\ =\frac{11.4mol}{\left(0.4564mol\right)+\left(11.4mol\right)}\\ =0.9615\end{array}$

Mole fraction of ethanol is $0.9615$

Vapour pressure of ethanol over the solution is calculated

$\begin{array}{c}{P}_{ethanol}=X_{ethanol}{P}^0{}_{ethanol}\\ =\left(0.9615\right)\times \left(760mmHg\right)\\ =730.7mmHg\end{array}$

Vapour pressure of ethanol over the solution is $730.7mmHg$

Interpretation Introduction

Interpretation: Mole fraction of DMG and molality of the solution are to be determined. Vapour pressure of ethanol over the solution and Boiling point of the solution are to be calculated.

Concept introduction:

Raoult’s law: In a solution, vapor pressure of solvent is proportional to its mole fraction.

  $P_{solvent}=X_{solvent}{P}^0{}_{solvent}$

  where,

  $P^0{}_{solvent}$ is the vapor pressure of pure solvent.

Mole fraction: Amount of that component divided by the total amount of all of the components of the mixture $\left({\text{ n}}_{\text{A}}{\text{ + n}}_{\text{B}}{\text{ + n}}_{\text{C }}\text{+ }\mathrm{...}\right)$.

  ${\text{Mole fraction of A (χ}}_{\text{A}}\text{)= }\frac{{\text{n}}_{\text{A}}}{{\text{ n}}_{\text{A}}{\text{ + n}}_{\text{B}}{\text{ + n}}_{\text{C }}\text{+ }\mathrm{...}}$

Answer to Problem 57GQ

Boiling point of the solution is $\text{79}{\text{.5}}^{\text{0}}\text{C}$

Explanation of Solution

(d)

Given,

  $\begin{array}{c}\text{Mass}\text{of}\text{ethanol}\text{=}\text{53}\text{.0}\text{g}\\ \text{Mass}\text{of}\text{DMG}\text{=}\text{525}\text{g}\\ \text{Molar}\text{mass}\text{of}\text{ethanol}\text{=}\text{46}\text{.07}\text{g/mol}\\ \text{Molar}\text{mass}\text{of}\text{DMG}\text{=}\text{116}\text{.12}\text{g/mol}\end{array}$

Number of moles of dimethylglyoxime is,

  $\begin{array}{c}{\text{n}}_{\text{DMG}}\text{=}\frac{\text{53}\text{.0}\text{g}}{\text{116}\text{.12}\text{g/mol}}\\ =0.4564mol\end{array}$

Molality of the solution is calculated

  $\begin{array}{c}\text{Molality (m) =}\frac{\text{0}\text{.4564}\text{mol}}{\text{0}\text{.525}\text{kg}}\\ \text{=}\text{0}\text{.869}\text{m}\end{array}$

Molality of the solution is $\text{0}\text{.869}\text{m}$

Elevation in boiling point of the solution is calculated

  $\begin{array}{c}{\text{ΔT}}_{\text{bp}}\text{=}{\text{K}}_{\text{bp}}{\text{m}}_{\text{solute}}\\ \text{=}\left(\text{1}{\text{.22}}^{\text{0}}\text{C/m}\right)\text{×}\left(\text{0}\text{.869}\text{m}\right)\\ =1.06{}^{\text{0}}\text{C}\end{array}$

Therefore,

Boiling point of the solution is $\left(\text{78}\text{.4}{}^{\text{0}}\text{C}\text{+}\text{1}\text{.06}{}^{\text{0}}\text{C}\right)\text{=}\text{79}{\text{.5}}^{\text{0}}\text{C}$