Question

Calculate ΔH
when 9.25×10−4 mol
of AgCl
dissolves in water.

Answer 1

`The enthalpy change (\(\Delta H\)) when a substance dissolves in water can be determined using the equation:`

`\(\Delta H = (q)/(n)\)`

`Where:- \(\Delta H\) = Enthalpy change (in joules/mol)- \(q\) = Heat absorbed or released (in joules)- \(n\) = Amount of substance (in moles)`

For the dissolution of AgCl in water, the enthalpy change is not provided. However, we can use a general value for the enthalpy of dissolution of AgCl.

`The enthalpy of dissolution of AgCl is approximately +65.5 kJ/mol.`

`First, convert this value to joules/mol:\(65.5 \text{ kJ/mol} = 65.5 * 10^3 \text{ J/mol}\)`

Given:

`\(n = 9.25 * 10^(-4) \text{ mol}\)`

Now, calculate the enthalpy change using the given formula:

`\(\Delta H = \frac{65.5 * 10^3 \text{ J/mol}}{1 \text{ mol}} * 9.25 * 10^(-4) \text{ mol}\)`

`\(\Delta H = 60.5375 \text{ J}\)`

`Therefore, when 9.25×10^−4 mol of AgCl dissolves in water, the enthalpy change (\(\Delta H\)) is approximately \(60.5375 \text{ J}\).`

Step-by-step explanation: