Chapter 12, Problem 12.83QP

Lysozyme is an enzyme that cleaves bacterial cell walls. A sample of lysozyme extracted from egg white has a molar mass of 13,930 g. A quantity of 0.100 g of this enzyme is dissolved in 150 g of water at 25°C. Calculate the vapor-pressure lowering, the depression in freezing point, the elevation in boiling point, and the osmotic pressure of this solution. (The vapor pressure of water at 25°C is 23.76 mmHg.)

Interpretation Introduction

Interpretation:

For given solution vapor pressure lowering, freezing point depression, boiling point elevation and osmotic pressures to be calculated.

Concept introduction

Boiling point elevation${\text{(ΔT}}_{\text{b}}\text{)}$: Boiling point is the distinction between boiling point of the pure solvent (${\text{T}}_{\text{b}}\text{°}$) and the boiling point of the solution (${\text{T}}_{\text{b}}$).

${\text{ΔT}}_{\text{b}}{\text{= K}}_{\text{b}}\text{m}$

Where,

$\begin{array}{l}{\text{ΔT}}_{\text{b}}\text{= Change ib boiling point}\\ {\text{T}}_{\text{b}}\text{= Boiling point of the solution}\\ {\text{T}}_{\text{b}}°\text{= Boiling point of pure solvent}\end{array}$

Freezing point depression${\text{(ΔT}}_{\text{f}})$: Freezing point depression is the distinction between freezing point of the pure solvent ${\text{(T}}_{\text{f}}\text{°)}$ and freezing point of the solution ${\text{(T}}_{\text{f}}\text{)}$.

${\text{ΔT}}_{\text{f}}{\text{= K}}_{\text{f}}\text{m}$

Where,

$\begin{array}{l}{\text{ΔT}}_{\text{f}}\text{= Change in freezing point}\\ {\text{T}}_{\text{f}}\text{= Freezing point of the solution}\\ {\text{T}}_{\text{f}}\text{°= Freezing point of pure solvent}\end{array}$

Osmotic pressure is the pressure that is needed to stop osmosis. Osmotic pressure of the solution is directly proportional to the concentration of the solution.  We can calculate osmotic pressure by using this formula is given by,

$\text{osmotic pressure(π) = MRT}$

Where,

Vapor pressure lowering: Vapor pressure lowering is one of the colligative properties. Pure solvent has higher vapour pressure than its solution have non-volatile liquid.  Thus vapour pressure lowering guide boiling point elevation.

${\text{ΔP = χ}}_{\text{B}}{\text{P°}}_{\text{A}}$

Where,

$\begin{array}{l}{\text{χ}}_{\text{B}}\text{- Mole fraction of the solute}\\ \text{P°C- vapor pressure of the pure solvent}\end{array}$

Answer to Problem 12.83QP

Vapour pressure lowering of the solution =

Freezing point elevation =

Boiling point elevation =

Osmotic pressure =

Explanation of Solution

Given data

Molar mass of egg white = $\text{13,930g}$

Amount of enzyme which is dissolved in water = $\text{0}\text{.100g}$

Amount of water = $\text{150g}$

Vapor pressure of water = $\text{23}\text{.76 mmHg at 25°C}$

Calculation of number of moles in lysozyme and water

${\text{n}}_{\text{Isozyme}}\text{=1}\text{.00g×}\frac{\text{1mol}}{\text{13,930g}}\text{=7}{\text{.18×10}}^{\text{-6}}\text{mol}$

Molecular mass of water =$\text{18}\text{.02g/mol}$

${\text{n}}_{\text{water}}\text{=150g×}\frac{\text{1mol}}{\text{18}\text{.02g}}\text{= 8}\text{.32mol}$

By plugging in the value of amount of Isozyme and molar mass of egg white, mole of Isozyme has calculated.  Similarly, by plugging in the value of amount of water and molar mass of water, mole of water has calculated.

Calculation of vapour pressure lowering of the solution

${\text{ΔP = χ}}_{\text{Iysozyme}}{\text{P°}}_{\text{water}}\text{=}\frac{{\text{n}}_{\text{Isozyme}}}{{\text{n}}_{\text{Isozyme}}\text{+}{\text{n}}_{\text{water}}}\text{(23}\text{.76mmHg)}$

$\text{ΔP=}\frac{\text{7}{\text{.18×10}}^{\text{-6}}\text{mol}}{\left[\text{(7}{\text{.18×10}}^{\text{-6}}\text{)+8}\text{.32mol}\right]}\text{(23}\text{.76mmHg) = 2}{\text{.05×10}}^{\text{-5}}\text{mmHg}$

By plugging in the values of mole fraction of Isozyme and vapour pressure of water, vapour pressure lowering of the solution has calculated.

Calculation freezing point depression of the solution

Molal freezing point depression constant = $\text{1}\text{.86°C/m}$

${\text{ΔT}}_{\text{f}}{\text{ =K}}_{\text{f}}\text{m =(1}\text{.86°C/m)}\left[\frac{\text{7}{\text{.18×10}}^{\text{-6}}\text{mol}}{\text{0}\text{.150kg}}\right]\text{= 8}{\text{.90×10}}^{\text{-5}}\text{°C}$

By plugging in the values of molal freezing point depression constant and molality of the solution, freezing point depression of the solution has calculated.

Calculation of boiling point elevation of the solution

Boiling point elevation constant = $\text{1}\text{.52°C/m}$

${\text{ΔT}}_{\text{b}}{\text{ =K}}_{\text{b}}\text{m =(1}\text{.86°C/m)}\left[\frac{\text{7}{\text{.18×10}}^{\text{-6}}\text{mol}}{\text{0}\text{.150kg}}\right]\text{= 2}{\text{.5×10}}^{\text{-5}}\text{°C}$

By plugging in the values of boiling point elevation constant and molality of the solution, boiling point elevation of the solution has calculated.

Calculation of osmotic pressure of the solution

As known above, we assume the density of the solution is $\text{1}\text{.00g/mL}\text{. The volume of the solution will be150mL}\text{.}$

$\text{π = MRT =}\left(\frac{\text{7}{\text{.18×10}}^{\text{-6}}\text{mol}}{\text{0}\text{.150L}}\right)\text{(0}\text{.08206L}\text{.atm/K}\text{.mol)(298K)=1}{\text{.17×10}}^{\text{-3}}\text{atm = 0}\text{.889mmHg}$

By plugging in the values of molarity of the solution, ideal gas constant and temperature in Kelvin, the osmotic pressure of the solution has calculated.

Conclusion

Vapour pressure lowering of the solution was calculated as $\text{2}{\text{.05×10}}^{\text{-5}}\text{mm Hg}$

Freezing point elevation has calculated as $\text{8}{\text{.90×10}}^{\text{-5}}\text{°C}\text{.}$

Boiling point elevation has calculated as $\text{2}{\text{.5×10}}^{\text{-5}}\text{°C}\text{.}$

Osmotic pressure has calculated as $\text{0}\text{.889 mmHg}\text{.}$