Answer:
[tex]\large \boxed{\text{1.36 mol/kg}}[/tex]
Explanation:
The formula for the boiling point elevation by an electrolyte is
[tex]\Delta T_{b} = iK_{b}b\\[/tex]
1. Calculate the boiling point elevation
ΔTb = 110.1 °C - 100.0 °C = 10.1 °C
2. Calculate the i-value
CoCl₃(aq) ⟶ Co³⁺(aq) + 3Cl⁻(aq)
One mole of CoCl₃ produces four moles of ions.
i = 4
3. Calculate the boiling point
[tex]\begin{array}{rcl}\\\Delta T_{\text{b}} &=& iK_{\text{b}}b\\10.1 \, ^{\circ}\text{C} & = & 4 \times 1.86 \, ^{\circ}\text{C} \cdot \text{kg}\cdot \text{mol}^{-1}\times b\\10.1 & = & 7.44b \text{ kg}\cdot \text{mol}^{-1} \\b & = & \dfrac{10.1}{\text{7.44 kg}\cdot \text{mol}^{-1} }\\\\ & = & \textbf{1.36 mol$\cdot$kg}^{\mathbf{-1}}\\\end{array}\\\text{The molal concentration of the cobalt(III) chloride is $\large \boxed{\textbf{1.36 mol$\cdot$kg}^{\mathbf{-1}}}$}[/tex]
