Question

Lysozyme extracted from chicken egg white has a molar mass of 13,930 g mo1-1. Exactly 0.1 g of this protein is dissolved in 50 g of water at 298 K. Calculate the vapor pressure lowering, the depression in freezing point, the elevation of boiling point, and the osmotic pressure of this solution. The vapor pressure of pure water at 298 K is 23. 76 mmHg.

Answer 1

Relative lowering in vapor pressure of solution is 23.7599 mmHg.

Depression in freezing point of the solution is 0.0002667 K.

Elevation in boiling point of the solution is 0.0000745 K.

Osmotic pressure of the solution is 0.0035 atm.

Step-by-step explanation:

1) Relative lowering in vapor pressure of solution containing non volatile solute is equal to the mole fraction of the solute.

`(p^o-p_s)/(p^o)=(n_2)/(n_1+n_2)`

`p^o`= Vapor Pressure of the pure solvent

`p_s`= Vapor Pressure of the solution

`n_2` moles of solute

`n_1` = moles of solvent

Given;
`p^o=23.76 mmHg ,p_s=?`

Moles of Lysozyme =
`n_2=(0.1 g)/(13,930 g/mol)=7.17* 10^(-6) mol`

Moles of water =
`n_1=(50 g)/(18 g/mol)=3.333 mol`

`(23.76 mmHg-p_s)/(23.76 mmHg)=(7.17* 10^(-6) mol)/(3.333 mol+7.17* 10^(-6) mol)`

`p_s=23.7599 mmHg`

2) Depression in freezing point
`\Delta T_f` is given by:

`\Delta T_f=K_f* m`

`K_f` = molal depression constant of solvent

m = molality of the solution =
`\frac{moles}{\text{mass of solvent in kg}}`

`\Delta T_f=1.86 K kg/mol* (7.17* 10^(-6) mol)/(0.050 kg)`

`\Delta T_f=0.0002667 K`

3) Elevation in boiling point
`\Delta T_b` is given by:

`\Delta T_b=K_b* m`

`K_b` = molal elevation constant of solvent

m = molality of the solution =
`\frac{moles}{\text{mass of solvent in kg}}`

`\Delta T_b=0.52 K kg/mol* (7.17* 10^(-6) mol)/(0.050 kg)`

`\Delta T_b=0.0000745 K`

4) Osmotic pressure of the solution
`\pi` is given as:

`\pi =cRT`

c = concentration of solution =
`(Moles)/(Volume(L))`

T = temperature of the solution

R = Universal gas constant = 0.0821 atm L/mol K

given , T = 298 K,

Mass of water = 50 g

Density of water = 1 g/ml

Volume of the water =
`(50 g)/(1 g/mL)=50 mL=0.050 L`

`c=(7.17* 10^(-6) mol)/(0.050 L)`

Since, there is less amount of solute in solvent volume of solution can be taken equal to the volume to the solution.

`\pi =(7.17* 10^(-6) mol)/(0.050 L)* 0.0821 atm L/mol K* 298 K`

`\pi=0.0035 atm`