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Solve the following.
An aqueous solution containing 12·50 g of barium chloride in 1000 g of water boils at 373·0834 K…
An aqueous solution containing $12\cdot50\text{ g}$ of barium chloride in $1000\text{ g}$ of water boils at $373\cdot0834\text{ K}$. Calculate the degree of dissociation of barium chloride.
Given $K_b$ for $\text{H}_2\text{O} = 0\cdot52\text{ K kg mol}^{-1}$; molecular mass of $\text{BaCl}_2 = 208\cdot34\text{ g mol}^{-1}$
Answer
Answer
Formula used: Elevation of boiling point: dTb = i*Kb*m with Kb = 0.52 K kg mol^-1, M2 = 208.34 g mol^-1, w1 = 1000 g, w2 = 12.5 g, dTb = 0.0834 K, so i = 2.673
$\Delta T_b = 373.0834 - 373.0000 = 0.0834\ \mathrm{K}$
$\Delta T_b = i K_b \dfrac{w_2 \times 1000}{M_2 \times w_1}$, so $0.0834 = i \times 0.52 \times \dfrac{12.5}{208.34}$, giving $i = 2.673$
$\mathrm{BaCl_2 \rightarrow Ba^{2+} + 2Cl^-}$ ($\nu = 3$), so $\alpha = \dfrac{i-1}{\nu-1} = \dfrac{2.673-1}{2} = 0.837$
Final answer: 0.837