Chapter 12, Problem 12.20QE

(a)

Interpretation Introduction

Interpretation:

The concentration of $\left[{\text{CO(NH}}_{\text{2}}{\text{)}}_{\text{2}}\right]$ in mass percentage has to be calculated.

Concept Introduction:

Mass percentage:  The mass percentage gives the number of grams of solute present in 100g of solution.  The formula of mass percentage is,

  $\text{Mass}\text{percentage=}\frac{\text{Grams}\text{of}\text{solute}}{\text{Grams}\text{of}\text{solution}}\text{×100\%}$

Answer to Problem 12.20QE

The concentration of $\left[{\text{CO(NH}}_{\text{2}}{\text{)}}_{\text{2}}\right]$ in mass percentage is $\text{2}\text{.9503\%}\text{.}$

Explanation of Solution

Given,

Mass of $\left[{\text{CO(NH}}_{\text{2}}{\text{)}}_{\text{2}}\right]$    =$\text{3}\text{.80 g}$

Mass of water        =$\text{125}\text{.0g}$

Mass of solution    =$\text{128}\text{.8g}$

The mass percentage of $\left[{\text{CO(NH}}_{\text{2}}{\text{)}}_{\text{2}}\right]$ is given as,

  $\begin{array}{l}\text{Mass}\text{percentage=}\frac{\text{Grams}\text{of}\text{solute}}{\text{Grams}\text{of}\text{solution}}\text{×100\%}\\ \\ \text{Mass}\text{percentage=}\frac{\text{3}\text{.80}\text{g}}{\text{128}\text{.8g}}\text{×100\%}\\ \\ \text{Mass}\text{percentage}\text{=2}\text{.9503\%}\end{array}$

The concentration of $\left[{\text{CO(NH}}_{\text{2}}{\text{)}}_{\text{2}}\right]$ in mass percentage is $\text{2}\text{.9503\%}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The concentration of $\left[{\text{CO(NH}}_{\text{2}}{\text{)}}_{\text{2}}\right]$ in mole fraction has to be calculated.

Concept Introduction:

Mole fraction:  Mole fraction of a substance in a solution is the number of moles of that substance divided by the total number of moles of all substances present.  The formula is,

  $\begin{array}{l}{\text{χ}}_{\text{A}}\text{=}\frac{\text{Moles}\text{of}\text{A (in}\text{mol)}}{\text{Moles}\text{of}\text{A (in mol) +Moles}\text{of}\text{B (in mol) +Moles}\text{of}\text{C (in mol) +}\mathrm{...}}\\ \\ {\text{χ}}_{\text{A}}\text{=}\frac{\text{Moles}\text{of}\text{A (in}\text{mol)}}{\text{Total}\text{number}\text{of}\text{moles}\text{of}\text{components}\text{(in}\text{mol)}}\end{array}$

Answer to Problem 12.20QE

The concentration of $\left[{\text{CO(NH}}_{\text{2}}{\text{)}}_{\text{2}}\right]$ in mole fraction is $\text{0}\text{.0090}\text{.}$

Explanation of Solution

Given,

Mass of $\left[{\text{CO(NH}}_{\text{2}}{\text{)}}_{\text{2}}\right]$    =$\text{3}\text{.80 g}$

Mass of water        =$\text{125}\text{.0g}$

Mass of solution    =$\text{128}\text{.8g}$

The moles of $\left[{\text{CO(NH}}_{\text{2}}{\text{)}}_{\text{2}}\right]$=$\text{3}\text{.80g×}\frac{\text{1}\text{mol}}{\text{60}\text{.06}\text{g}}\text{=0}\text{.0632}\text{moles}$

The moles of water=$\text{125g×}\frac{\text{1}\text{mol}}{\text{18}\text{.02}\text{g}}\text{=6}\text{.9367moles}$

The mole fraction of $\left[{\text{CO(NH}}_{\text{2}}{\text{)}}_{\text{2}}\right]$,

  $\begin{array}{l}{\text{χ}}_{\left[{\text{CO(NH}}_{\text{2}}{\text{)}}_{\text{2}}\right]}\text{=}\frac{\text{Moles}\text{of}\left[{\text{CO(NH}}_{\text{2}}{\text{)}}_{\text{2}}\right]\text{ (in}\text{mol)}}{\text{Total}\text{number}\text{of}\text{moles}\text{of}\text{components}\text{(in}\text{mol)}}\\ \\ {\text{χ}}_{\left[{\text{CO(NH}}_{\text{2}}{\text{)}}_{\text{2}}\right]}\text{=}\frac{\text{0}\text{.0632}\text{moles}}{\text{0}\text{.0632}\text{moles+ 6}\text{.9367}\text{moles}}\\ \\ {\text{χ}}_{\left[{\text{CO(NH}}_{\text{2}}{\text{)}}_{\text{2}}\right]}\text{=}\frac{\text{0}\text{.0632}\text{moles}}{\text{6}\text{.9999}\text{moles}}\\ \\ {\text{χ}}_{\left[{\text{CO(NH}}_{\text{2}}{\text{)}}_{\text{2}}\right]}\text{=0}\text{.0090}\end{array}$

The concentration of $\left[{\text{CO(NH}}_{\text{2}}{\text{)}}_{\text{2}}\right]$ in mole fraction is $\text{0}\text{.0090}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The concentration of $\left[{\text{CO(NH}}_{\text{2}}{\text{)}}_{\text{2}}\right]$ in molality has to be calculated.

Concept Introduction:

Molality:  Molality is defined as moles of solute per specified amount of mass of the solvent.  The formula is,

  $\text{Molality}\text{(m)=}\frac{\text{Mass of}\text{solute (in mol)}}{\text{Mass}\text{of}\text{solvent (in kg)}}$

Answer to Problem 12.20QE

The concentration of $\left[{\text{CO(NH}}_{\text{2}}{\text{)}}_{\text{2}}\right]$ in molality is $\text{0}\text{.51m}\text{.}$

Explanation of Solution

Given,

Mass of $\left[{\text{CO(NH}}_{\text{2}}{\text{)}}_{\text{2}}\right]$    =$\text{3}\text{.80 g}$

Mass of water        =$\text{125}\text{.0g=0}\text{.125}\text{kg}$

Mass of solution    =$\text{128}\text{.8g}$

The moles of $\left[{\text{CO(NH}}_{\text{2}}{\text{)}}_{\text{2}}\right]$=$\text{3}\text{.80g×}\frac{\text{1}\text{mol}}{\text{60}\text{.06}\text{g}}\text{=0}\text{.0632}\text{moles}$

The concentration of $\left[{\text{CO(NH}}_{\text{2}}{\text{)}}_{\text{2}}\right]$ is,

  $\begin{array}{l}\text{Molality}\text{(m)=}\frac{\text{Mass of}\text{solute (in mol)}}{\text{Mass}\text{of}\text{solvent (in kg)}}\\ \\ \text{m=}\frac{\text{0}\text{.0632}\text{moles}}{\text{0}\text{.125}\text{kg}}\\ \\ \text{m=0}\text{.5056}\text{m}\approx \text{0}\text{.51m}\end{array}$

The concentration of $\left[{\text{CO(NH}}_{\text{2}}{\text{)}}_{\text{2}}\right]$ in molality is $\text{0}\text{.51m}\text{.}$