Answer:
the answer would be a i did the test
Explanation:
Answer:
Option d.
1 mole AlCl3in 500 g water
Explanation:
ΔT = Kf . m . i
Freezing T° of solution = - (Kf . m . i)
In order to have the lowest freezing T° of solution, we need to know which solution has the highest value for the product (Kf . m . i)
Kf is a constant, so stays the same and m stays also the same because we have the same moles, in the same amount of solvent. In conclussion, same molality to all.
i defines everything. The i refers to the Van't Hoff factor which are the number of ions dissolved in solution. We assume 100 & of ionization so:
a. Glucose → i = 1
Glucose is non electrolytic, no ions formed
b. MgF₂ → Mg²⁺ + 2F⁻
i = 3. 1 mol of magnessium cation and 2 fluorides.
c. KBr → K⁺ + Br⁻
i = 2. 1 mol potassium cation and 1 mol of bromide anion
d. AlCl₃ → Al³⁺ + 3Cl⁻
i = 4. 1 mol of aluminum cation and 3 mol of chlorides.
Kf . m . 4 → option d will has the highest product, therefore will be the lowest freezing point.
Answer: 
Explanation:
cM 0 0
So dissociation constant will be:
Given: c = 0.15 M
pH = 1.86
= ?
Putting in the values we get:
Also ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
![1.86=-log[H^+]](https://tex.z-dn.net/?f=1.86%3D-log%5BH%5E%2B%5D)
![[H^+]=0.01](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.01)
![[H^+]=c\times \alpha](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Ctimes%20%5Calpha)
As ![[H^+]=[ClCH_2COO^-]=0.01](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5BClCH_2COO%5E-%5D%3D0.01)
![K_a=1.67\times 10^{-3]](https://tex.z-dn.net/?f=K_a%3D1.67%5Ctimes%2010%5E%7B-3%5D)
Thus the vale of for the acid is
