Question

A 4.00 m solution of KCl was prepared using 1.00 kg of water at 25.0 ∘C. Once the solid had all dissolved, the temperature of the solution was 12.3 ∘C. Calculate the heat of solution, ΔHsoln, of KCl. Assume that the specific heat of the solution is identical to that of water, 4.18 J/(g⋅∘C).

Answer 1

Answer:

The heat of solution of KCl is 17.2 kJ

Explanation:

Step 1: given data

We have a 4m solution of KCl prepared in 1kg of water

⇒molality = number of moles / kg of solute

⇒ in this case it means 4 moles / 1kg of water = 4

Specific heat of water = 4.18J/g °C

KCl has a molar mass of 75.55g/mole

Step 2: calculate mass and change of temperature

Heat of solution = mass (g) * specific heat (j/g °C) * change of temperature

mass of KCl we can calculate through te formula mass = moles * Molar mass

⇒ m(KCl) = 4 moles * 74.55g/mole = 298.2g

⇒ total mass = 298.2g of KCl + 1000g of water = 1298.2g

Change of temperature = Final temperature T2 - initial temperature T1 = 12.3°C - 25°C = -12.7 °C   ⇔ this means the temperature reduces with 12.7 °C

Step3 : calculating heat of solution

heat of solution (q) = mass * specific heat * change in temperature

q = (1298g) * (4.18J/g °C ) * (-12.7) = -68910 J = -68.910 kJ

We will use the positve value of this

ΔH = q / moles = 68.91 kJ / 4 moles = 17.2 kJ

The heat of solution of KCl is 17.2 kJ