Answer:
6.82g
0.59moles
Explanation:
1. What is the mass sample of 0.0500 moles of zinc chloride ?
Given parameters:
Number of moles ZnCl₂ = 0.05moles
Unknown:
Mass of the sample = ?
Solution:
To find the mass of a substance using the number of moles, it would be pertinent to understand what mole is.
A mole is a substance that contains the avogadro's number of particles.
It relates to the mass using the expression below;
Mass of a substance = number of moles x molar mass
Molar mass of ZnCl₂;
Atomic mass of Zn = 65.4g/mol
Cl = 35.5g/mol
Molar mass = 65.4 + 2(35.5) = 136.4g/mole
Mass of a substance = 0.05 x 136.4 = 6.82g
2. How many moles of potassium sulfide are in a 65.50g sample?
Given parameters:
Mass of K₂S = 65.5g
Unknown:
Number of moles = ?
Solution:
The number of moles of any substance is related to mass using the expression below;
Number of moles = 
Molar mass of K₂S = 2(39) + 32 = 110g/mol
Number of moles =
= 0.59moles
Answer:
The degree of dissociation of acetic acid is 0.08448.
The pH of the solution is 3.72.
Explanation:
The 
The value of the dissociation constant = 
![pK_a=-\log[K_a]](https://tex.z-dn.net/?f=pK_a%3D-%5Clog%5BK_a%5D)

Initial concentration of the acetic acid = [HAc] =c = 0.00225
Degree of dissociation = α

Initially
c
At equilibrium ;
(c-cα) cα cα
The expression of dissociation constant is given as:
![K_a=\frac{[H^+][Ac^-]}{[HAc]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH%5E%2B%5D%5BAc%5E-%5D%7D%7B%5BHAc%5D%7D)



Solving for α:
α = 0.08448
The degree of dissociation of acetic acid is 0.08448.
![[H^+]=c\alpha = 0.00225M\times 0.08448=0.0001901 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Calpha%20%3D%200.00225M%5Ctimes%200.08448%3D0.0001901%20M)
The pH of the solution ;
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![=-\log[0.0001901 M]=3.72](https://tex.z-dn.net/?f=%3D-%5Clog%5B0.0001901%20M%5D%3D3.72)
Answer:
173.83 mmHg is the vapor pressure of a ethylene glycol solution.
Explanation:
Vapor pressure of water at 65 °C=
Vapor pressure of the solution at 65 °C= 
The relative lowering of vapor pressure of solution in which non volatile solute is dissolved is equal to mole fraction of solute in the solution.
Mass of ethylene glycol = 22.37 g
Mass of water in a solution = 82.21 g
Moles of water=
Moles of ethylene glycol=



173.83 mmHg is the vapor pressure of a ethylene glycol solution.