Question

When 5.325 g of KCl dissolved in 100.0 ML of water, the temperature of the solution decreases from 24.1 Celsius to 21.2 Celsius. Calculate the amount of heat required for the KCl to dissolve.

Calculate per ?H mol of KCl

Answer 1

First calculate mass of solution

Mass of solution (m) = Mass of water + Mass of KCl

Assume density of water = 1.00 g /ml. Therefore, Mass of water = density x volume = 1.00 g / ml x 100 ml = 100.0 g

Therefore, Mass of solution = 100 g + 5.325 g = 105.325 g

Specific heat capacity of solution ( C) = 4.18 J / g ⁰ C

Change in temperature ( T) of solution = T final - T initial = 21.2 ⁰ C - 24.1 ⁰ C = - 2.9 ⁰ C

In this case, q reaction = - q solution

Hence ,Enthalpy change of solution =- m x C x   T

Enthalpy change of solution = - 105.325 g x 4.18 J / g ⁰ C x ( - 2.9 ⁰ C )

= 1276.7 J

Enthalpy change of solution on dissolution of 1 g KCl = 1276.7 J / 5.325 g =239.8 J / g

Molar mass of KCl = 39.10 + 35.45 = 74.55 g /mol

Enthalpy change of solution on dissolution of 1 mole KCl =239.8  J / g x 74.55 g /mol = 17877.1 J / mol

Enthalpy change of solution on dissolution of 1 mole KCl = 17877.1 J / mol = 17.877 kJ / mol