Respuesta :
Solution :
When non volatile solute is added to solvent, vapor pressure gets lowered.
Relative lowering in vapor pressure is given :
[tex]$\frac{P^0-P}{P^0}$[/tex] = [tex]$\text{mole fraction}$[/tex] of solute
[tex]$\frac{P^0-P}{P^0}=x_B$[/tex]
[tex]$P^0$[/tex] = vapor pressure of pure solvent
P = vapor pressure of solution
[tex]$x_B$[/tex] = mole fraction of solute
[tex]$x_B=\frac{n_B}{n_A+n_B}$[/tex]
[tex]$n_B $[/tex] = [tex]$\text{number of moles of solute}$[/tex]
[tex]$n_A$[/tex] = [tex]$\text{number of moles of solvent}$[/tex]
Number of moles [tex]$=\frac{\text{weight}}{\text{molecular weight}}$[/tex]
[tex]$\frac{P^0-P}{P^0}=\frac{w_B/M_B}{w_A/M_A+w_B/M_B}$[/tex]
[tex]$\approx \frac{w_B/M_B}{w_A/M_A}$[/tex]
1. For 10 g of [tex]$CH_3COOK$[/tex]
[tex]$CH_3COOK \rightarrow CH_3COO^- + K^+$[/tex]
Ions = 2
It will affect colligative property.
[tex]$\frac{P^0-P}{P^0} = \frac{i \times 10/98}{w_A/M_A}$[/tex]
Relative lowering in vapor pressure will be :
[tex]$=\frac{2 \times 10/98}{w_A/M_A}$[/tex]
[tex]$=\frac{0.20}{w_A/M_A}$[/tex]
2. For 20 g sucrose
Sucrose is non electrolyte, i = 1
[tex]$\frac{P^0-P}{P^0} = \frac{ 20/342}{w_A/M_A}$[/tex]
[tex]$=\frac{0.050}{w_A/M_A}$[/tex]
3. For 20 g of glucose.
Glucose is a non electrolyte, i = 1
[tex]$\frac{P^0-P}{P^0} = \frac{20/180}{w_A/M_A}$[/tex]
[tex]$=\frac{0.11}{w_A/M_A}$[/tex]
[tex]$w_A/M_A$[/tex] is same in all three solutions.
Hence, lowering in vapor pressure is maximum in [tex]$CH_3COOK$[/tex] and minimum is Sucrose.
Vapor pressure from lowest to highest.
10 g of [tex]$CH_3COOK$[/tex] < 20 g of glucose < 20 g of sucrose
The pressure exerted by vapor to the other gas is called vapor pressure.
The formula used to solve the question is as follows:-
[tex]\frac{P^o -P}{P^o}[/tex]
The water vapor depends on the following:-
- Water pressure
- Temperature.
After solving the equation, the correct sequence is as follows:-
[tex]CH_3COOK > GLUCOSE > SUCROSE[/tex]
For more information, refer to the link:-
https://brainly.com/question/22810476
