You are attempting to determine the thermodynamic properties of an unknown compound. You’ve already determined delta H fusion to be 5.56 kJ/mol and delta H vaporization to be 21.99 kJ/mol. In one of these previous experiments, 2.56 g of this compound was placed in a glass bulb and vaporized by placing the glass bulb into 200. g of hot water initially at 68.25 degrees Celsius, which resulted in a final water temperature of 67.91 degrees Celsius. You are now attempting to determine the enthalpy of solution of this compound. To this end, you dissolve 55.6 g of this compound in 495 g of water initially at 25.2 degrees Celsius. The final temperature of the solution is 19.4 degrees Celsius. What is the enthalpy of solution of this compound in kJ/mol?

The heat of solution of the compound = 42.75 kJ/mol

Explanation:

Here we have;

The heat of vaporization = 21.99 kJ/mol

2.56 g of the compound is then placed in 200 g. of water

The temperature of the water reduces from 68.25°C to 67.91°C

The specific heat capacity of water = 4.19 kJ/kg = 4.19 J/g

Mass of the water = 200 g

Heat lost by the water ΔH = m × c × Δθ = n × ΔH vaporization of the compound

Where;

m = The mass of the water = 200 g

c = The specific heat capacity of water = 4.19 kJ/kg = 4.19 J/g

Δθ = The change in temperature = 68.25°C - 67.91°C = 0.34°C

n = The number of moles of the compound

ΔH = m × c × Δθ = 200×4.19×0.34 = 284.92 J = 0.28492 kJ

∴ n × ΔH vaporization of the compound = 0.28492 kJ

n = 0.28492 kJ/( ΔH vaporization of the compound) = 0.28492/21.99≈0.013 moles

∴ 2.56 g = 0.013 moles

1 mole = The molar mass of the compound = 2.56/0.013= 197.6 g/mol

Therefore;

The number of moles in 55.6 g of the compound = 55.6/197.6 = 0.2814 moles

The heat given by the water = ΔH = 495 × (25.2 - 19.4) × 4.19 = 12029.49 J

Therefore;

0.2814 × ΔH(solution) = 12029.49 J = 12.02949 kJ

ΔH(solution) = 12.02949/0.2814 = 42.75 kJ/mol

The heat of solution of the compound = 42.75 kJ/mol.

it means there are 3 atoms of oxygen

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