Ksp of PbBr₂ is 6.60 × 10⁻⁶. The molar solubility of PbBr₂ in pure water is 0.0118M.
Ksp or Solubility Product Constant is an equilibrium constant for the dissociation in an aqueous solution.
Molar solubility (S) is the concentration of the dissolved substance in a solution that is saturated.
Let the molar solubility be S upon dissociation.
PbBr₂ or Lead Bromide dissociates in pure water as follows:
PbBr₂ ----------> Pb⁺² + Br⁻
S 2S
Ksp = [Pb⁺²] [ Br⁻]
Ksp = (S) (2S)²
Ksp = 4S³
6.60 × 10⁻⁶ = 4S³
S = 0.0118M
Hence, the Molar solubility S is 0.0118M.
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Yes, anything that lives. Cells have a nucleus, they are alive and can reproduce. (Not man made/abiotic)
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
