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An aqueous solution containing 12·48 g of barium chloride (BaCl₂) in 1000 g of water, boils at…
An aqueous solution containing $12\cdot48\text{ g}$ of barium chloride ($\text{BaCl}_2$) in $1000\text{ g}$ of water, boils at $100\cdot0832^\circ\text{C}$. Calculate the degree of dissociation of barium chloride. ($K_b$ for water = $0\cdot52\text{ K kg mol}^{-1}$, at. wt. $\text{Ba} = 137$, $\text{Cl} = 35\cdot5$)
Answer
Answer
Formulas used: Elevation of boiling point: dTb = i*Kb*m with Kb = 0.52 K kg mol^-1, M2 = 208 g mol^-1, w1 = 1000 g, w2 = 12.48 g, dTb = 0.0832 K, so i = 2.667 Degree of dissociation from i: alpha = (i - 1)/(nu - 1) with i = 2.667, nu = 3, so alpha = 0.8333
$\Delta T_b = 100\cdot0832 - 100 = 0\cdot0832\text{ K}$; $M_2(\mathrm{BaCl_2}) = 137 + 2\times35\cdot5 = 208\text{ g mol}^{-1}$
$\Delta T_b = iK_b\dfrac{w_2\times1000}{M_2\times w_1}$, so $0\cdot0832 = i\times0\cdot52\times\dfrac{12\cdot48}{208\times1}$
$i = 2\cdot67$
$\mathrm{BaCl_2 \rightarrow Ba^{2+} + 2Cl^-}$, so $\nu = 3$ and $\alpha = \dfrac{i-1}{\nu-1} = \dfrac{2\cdot67-1}{2} = 0\cdot833$
Final answer: 0.833