Answered

3.15 mol of an unknown solid is placed into enough water to make 150.0 mL of solution. The solution's temperature increases by 19.2°C. Calculate ∆H for the dissolution of the unknown solid. (The specific heat of the solution is 4.18 J/g･°C and the density of the solution is 1.20 g/mL).

Answer :

The heat generated following the dissolution of the unknown solid in

water is used to find its heat of dissolution.

The heat of dissolution of the unknown solid is approximately 4.586 kJ/mol

Reasons:

The given parameter are;

Number of moles of the unknown solid = 3.15 moles

The volume of the resulting solution = 150.0 mL

The temperature change, ΔT  = 19.2 °C

Density of solution = 1.20 g/mL

Heat capacity of the solution, c = 4.18 J/(g·°C)

Required:

The ΔH for the reaction.

Solution:

Assumption; The heat capacity of the resulting solution is equal to the heat

capacity of water.

The heat gained to raise 150.0 mL of the solution by 19.2 °C is given as

follows;

Mass of the solution, m = 150.0 mL × 1.20 g/mL = 180 g

Heat gained, Q = m·c·ΔT

The heat gained by the solution, Q = 180 × 4.18 × 19.2 = 14446.08

∴ The heat gained by the solution, Q = 14,446.08 J

The heat of dissolution = Heat change ÷ Number of moles of substance

Therefore;

[tex]Heat \ of \ dissolution = \dfrac{14,446.08 \ J}{3.15 \ mol} \approx 4586.06 \ J/mol \approx 4.586 \ kJ/mol[/tex]

  • The heat of dissolution of the unknown solid, [tex]\Delta H_{diss}[/tex] ≈ 4.586 kJ/mol

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