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

Answer:

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.

Explanation:

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

$\frac{p^o-p_s}{p^o}=\frac{n_2}{n_1+n_2}$

$p^o$= 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 =$n_2=\frac{0.1 g}{13,930 g/mol}=7.17\times 10^{-6} mol$

Moles of water =$n_1=\frac{50 g}{18 g/mol}=3.333 mol$

$\frac{23.76 mmHg-p_s}{23.76 mmHg}=\frac{7.17\times 10^{-6} mol}{3.333 mol+7.17\times 10^{-6} mol}$

$p_s=23.7599 mmHg$

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

$\Delta T_f=K_f\times m$

$K_f$ = molal depression constant of solvent

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

$\Delta T_f=1.86 K kg/mol\times \frac{7.17\times 10^{-6} mol}{0.050 kg}$

$\Delta T_f=0.0002667 K$

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

$\Delta T_b=K_b\times m$

$K_b$ = molal elevation constant of solvent

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

$\Delta T_b=0.52 K kg/mol\times \frac{7.17\times 10^{-6} mol}{0.050 kg}$

$\Delta T_b=0.0000745 K$

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

$\pi =cRT$

c = concentration of solution =$\frac{Moles}{Volume(L)}$

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 = $\frac{50 g}{1 g/mL}=50 mL=0.050 L$

$c=\frac{7.17\times 10^{-6} mol}{0.050 L}$

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

$\pi =\frac{7.17\times 10^{-6} mol}{0.050 L}\times 0.0821 atm L/mol K\times 298 K$

$\pi=0.0035 atm$