Question

0.391 mol of a solid was dissolved in 330 mL of water at 22.5 ^oC. After the solid had fully dissolved, the final temperature of the solution was 25.7 ^oC. What is the molar heat of solution of the substance?

Answer 1

Solution:-

​​​​​​q = m × C × ∆T

Where, q = heat gained by water

m = mass of water

C = specific heat of water

∆T = change in temperature.

mass of water (m) = 330 g  

(Density of water = 1g/ml , volume given = 330ml , thus, mass = density × volume = (1 g/ml × 330ml) = 330g )

C = 4.18 J/g -°C

∆T = (25.7 - 22.5) = 3.2°C

q = 330 × 4.18 × 3.2 = 4414.08 J

As heat absorbed by water is released by dissolution of solid in water, so we have to take negative value of heat absorbed by water during calculation of molar heat of solution (∆H).

∆H = -(q) / n

where, n = moles of solid (solute) = 0.391 mol

Thus, Molar Heat of solution of the substance,

∆H = -4414.08 / 0.391 = -11289.2 J/mol = -11.2892 kJ/mol

----------------------------------------------------------------