Answer:
Boiling point of the solution is 100.78°C
Explanation:
This is about colligative properties.
First of all, we need to calculate molality from the freezing point depression.
ΔT = Kf . m . i
As the solute is nonelectrolyte, i = 1
0°C - (-2.79°C) = 1.86 °C/m . m . 1
2.79°C / 1.86 m/°C = 1.5 m
Now, we go to the boiling point elevation
ΔT = Kb . m . i
Final T° - 100°C = 0.52 °C/m . 1.5m . 1
Final T° = 0.52 °C/m . 1.5m . 1 + 100°C → 100.78°C
Answer:
mass HF = 150.05 g
Explanation:
- SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(l)
⇒ Q = (ΔH°rxn * mHF) / (mol HF * MwHF )
∴ MwHF = 20.0063 g/mol
∴ mol HF = 4 mol
∴ ΔH°rxn = - 184 KJ
∴ Q = 345 KJ
mass HF ( mHF ):
⇒ mHF = ( Q * mol HF * MwHF ) / ΔH°rxn
⇒ mHF = ( 345 KJ * 4mol HF * 20.0063 g/mol ) / 184 KJ
⇒ mHF = 150.05 g HF
5. 1 hydrogen, 1 Nitrogen, 3 oxygen
