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Solve the following.
An aqueous solution containing 12·48 g of barium chloride in 1000 g of water boils at 373·0832 K…
An aqueous solution containing $12\cdot48\text{ g}$ of barium chloride in $1000\text{ g}$ of water boils at $373\cdot0832\text{ K}$. Calculate the degree of dissociation ($\alpha$) of barium chloride.
$K_b$ 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
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 = 1000 g, w2 = 12.48 g, dTb = 0.0832 K, so i = 2.671 Degree of dissociation from i: alpha = (i - 1)/(nu - 1) with i = 2.671, nu = 3, so alpha = 0.836
Formula: $\Delta T_b = i K_b \dfrac{w_2 \times 1000}{M_2 \times w_1}$ and $\alpha = \dfrac{i - 1}{n - 1}$
Substitution:
$\Delta T_b = 373\cdot0832\text{ K} - 373\text{ K} = 0\cdot0832\text{ K}$
$0\cdot0832 = i \times 0\cdot52\text{ K kg mol}^{-1} \times \dfrac{12\cdot48\text{ g}}{208\cdot34\text{ g mol}^{-1} \times 1\text{ kg}}$
$i = 2\cdot671$
$\mathrm{BaCl_2 \rightarrow Ba^{2+} + 2Cl^-}$, so $n = 3$
$\alpha = \dfrac{2\cdot671 - 1}{3 - 1} = 0\cdot836$
Final answer: 0.836