Answer:
[tex]\large \boxed{\text{37.0 kJ$\cdot$mol$^{-1}$}}[/tex]
Explanation:
There are two heat transfers involved in this problem.
Heat of solution of AlCl₃ + heat lost by water = 0
q₁ + q₂ = 0
nΔH + mCΔT = 0
Let's calculate the heats separately.
[tex]\text{Moles of NH$_{4}$Cl}= \text{3.00 g NH$_{4}$Cl} \times \dfrac{\text{1 mol NH$_{4}$Cl}}{\text{133.34 g NH$_{4}$Cl}} = \text{0.022 50 mol NH$_{4}$Cl}\\\\q_{1} = \text{0.022 50 mol} \times \Delta H[/tex]
m = 100.00 g
C = 4.18 J·°C⁻¹g⁻¹
ΔT = -1.99 °C
q₂ = mCΔT = 100.00 g × 4.184 J·°C⁻¹g⁻¹ × (-1.99 °C) = -832.6 J
[tex]q_{1} + q_{2} = \text{0.022 50 mol} \times \Delta H - \text{832.6 J} = 0\\\text{0.022 50 mol} \times \Delta H = \text{832.6 J}\\\Delta H = \dfrac{\text{832.6 J}}{\text{0.022 50 mol}} = \text{37 010 J/mol} =\textbf{37.0 kJ}\cdot\textbf{mol}^{\mathbf{-1}}\\\\\text{The heat of solution of NH$_{4}$Cl is $\large \boxed{\textbf{37.0 kJ$\cdot$mol$^{\mathbf{-1}}$}}$}[/tex]
