Answer:
[tex]\large \boxed{\text{7.7 kJ}\cdot\text{mol}^{-1} }[/tex]
Explanation:
1. Mass of solution
[tex]\text{Mass of solution} = \text{150 mL} \times \dfrac{\text{1.02 g}}{\text{1 mL}} = \text{153 g}[/tex]
2. Calorimetry
There are two energy flows in this reaction.
q₁ = heat from reaction
q₂ = heat to warm the water
[tex]\begin{array}{ccccl}n\Delta H & + & mC\Delta T &=&0\\\text{0.70 mol}\times \Delta H& + & \text{153 g} \times 4.18 \text{ J}\cdot^{\circ}\text{C}^{-1}\cdot \text{g⁻1} \times (-8.4\, ^{\circ}\text{C}) & = & 0\\0.70\Delta H \text{ mol} & - & \text{5372 J} & = & 0\\& & 0.70\Delta H \text{ mol} & = & \text{5372 J} \\\end{array}[/tex]
[tex]\begin{array}{ccrcl} & & \Delta H & = & \dfrac{\text{ 5372 J}}{\text{0.70 mol}}\\\\ & & & = & \text{7670 J/mol}\\ & & & = & \textbf{7.7 kJ}\cdot\textbf{mol}^{ -\mathbf{-1}} \\\end{array}\\\text{The heat of solution of the compound is $\large \boxed{\textbf{7.7 kJ}\cdot\textbf{mol}^{\mathbf{-1}}}$}[/tex]
The heat of solution is positive, because the water cooled down when the substance dissolved,
