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Answer the following.
An aqueous solution containing 1·25 g BaCl₂ (molecular mass 208·34 g mol⁻¹) in 100 g of water boils…
An aqueous solution containing $1\cdot25\text{ g}$ $\text{BaCl}_2$ (molecular mass $208\cdot34\text{ g mol}^{-1}$) in $100\text{ g}$ of water boils at $100\cdot085^\circ\text{C}$.
Calculate the degree of dissociation of $\text{BaCl}_2$.
($K_b$ for water $= 0\cdot52\text{ K kg mol}^{-1}$)
Answer
Answer
Formulas used: Elevation of boiling point: dTb = i*Kb*m with Kb = 0.52 K kg mol^-1, M2 = 208.34 g mol^-1, w1 = 100 g, w2 = 1.25 g, dTb = 0.085 K, so i = 2.72 Degree of dissociation from i: alpha = (i - 1)/(nu - 1) with i = 2.72, nu = 3.0, so alpha = 0.86
$\Delta T_b = 100.085 - 100 = 0.085\ \mathrm{K}$; $m = \dfrac{1.25/208.34}{0.100} = 0.0600\ \mathrm{mol\ kg^{-1}}$
$\Delta T_b = i K_b m \Rightarrow i = \dfrac{0.085}{0.52 \times 0.06} = 2.72$
$\mathrm{BaCl_2 \rightarrow Ba^{2+} + 2Cl^-}$ ($\nu = 3$): $\alpha = \dfrac{i - 1}{\nu - 1} = \dfrac{2.72 - 1}{2} = 0.862$
Final answer: 0.862