Answer:
4552 mL
Explanation:
From the question given above, the following data were obtained:
Volume of stock solution (V₁) = 55 mL
Molarity of stock solution (M₁) = 12 M
Molarity of diluted solution (M₂) = 0.145 M
Volume of diluted solution (V₂) =?
The volume of the diluted solution can be obtained by using the dilution formula as illustrated below:
M₁V₁ = M₂V₂
12 × 55 = 0.145 × V₂
660 = 0.145 × V₂
Divide both side by 0.145
V₂ = 660 / 0.145
V₂ ≈ 4552 mL
Thus, the volume of the diluted solution is 4552 mL
For the purpose, we will use the equation for determining the dissociation constant from concentration and <span>percent of ionization:
Kd = c </span>× α²
α = √(Kd/c) × 100%
Kd = 6.0×10⁻⁷
c(HA) = 0.1M
α = √(6.0×10⁻⁷/0.1) × 100% = 0.23%
So, in the solution, the acid <span>percent of ionization will be just 0.23%.</span>
Answer:
Sorry how no this equation guys
13.82 g / 180.16 g/mol = .07671 moles
.07671 moles / (86.18 g / 1000 g/kg) = .8901 molal
Let me know if you have any further questions!