Answer:
Molar mass→ 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol
Explanation:
Let's apply the formula for freezing point depression:
ΔT = Kf . m
ΔT = 74.2°C - 73.4°C → 0.8°C
Difference between the freezing T° of pure solvent and freezing T° of solution
Kf = Cryoscopic constant → 5.5°C/m
So, if we replace in the formula
ΔT = Kf . m → ΔT / Kf = m
0.8°C / 5.5 m/°C = m → 0.0516 mol/kg
These are the moles in 1 kg of solvent so let's find out the moles in our mass of solvent which is 0.125 kg
0.0516 mol/kg . 0.125 kg = 6.45×10⁻³ moles. Now we can determine the molar mass:
Molar mass (mol/kg) → 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol
Answer: (a) The solubility of CuCl in pure water is
.
(b) The solubility of CuCl in 0.1 M NaCl is
.
Explanation:
(a) Chemical equation for the given reaction in pure water is as follows.

Initial: 0 0
Change: +x +x
Equilibm: x x

And, equilibrium expression is as follows.
![K_{sp} = [Cu^{+}][Cl^{-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BCu%5E%7B%2B%7D%5D%5BCl%5E%7B-%7D%5D)

x = 
Hence, the solubility of CuCl in pure water is
.
(b) When NaCl is 0.1 M,
, 
, 
Net equation: 
= 0.1044
So for, 
Initial: 0.1 0
Change: -x +x
Equilibm: 0.1 - x x
Now, the equilibrium expression is as follows.
K' = 
0.1044 = 
x = 
Therefore, the solubility of CuCl in 0.1 M NaCl is
.