Answer:
127.3° C, (This is not a choice)
Explanation:
This is about the colligative property of boiling point.
ΔT = Kb . m . i
Where:
ΔT = T° boling of solution - T° boiling of pure solvent
Kb = Boiling constant
m = molal (mol/kg)
i = Van't Hoff factor (number of particles dissolved in solution)
Water is not a ionic compound, but we assume that i = 2
H₂O → H⁺ + OH⁻
T° boling of solution - 118.1°C = 0.52°C . m . 2
Mass of solvent = Solvent volume / Solvent density
Mass of solvent = 500 mL / 1.049g/mL → 476.6 g
Mol of water are mass / molar mass
76 g / 18g/m = 4.22 moles
These moles are in 476.6 g
Mol / kg = molal → 4.22 m / 0.4766 kg = 8.85 m
T° boling of solution = 0.52°C . 8.85 m . 2 + 118.1°C = 127.3°C
Answer:
Molar heat of solution of KBr is 20.0kJ/mol
Explanation:
Molar heat of solution is defined as the energy released (negative) or absorbed (Positive) per mole of solute being dissolved in solvent.
The dissolution of KBr is:
KBr → K⁺ + Br⁻
In the calorimeter, the temperature decreases 0.370K, that means the solution absorbes energy in this process. The energy is:
q = 1.36kJK⁻¹ × 0.370K
q = 0.5032kJ
Moles of KBr in 3.00g are:
3.00g × (1mol / 119g) = 0.0252moles
Thus, molar heat of solution of KBr is:
0.5032kJ / 0.0252moles = <em>20.0kJ/mol</em>
Answer:
NH4)3PO4
this is a filler: jsjdkflflflsojsskksdnfkdod
because it a essential for good heat transfer
