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Solve the following.
A solution containing 0·5 g of KCl dissolves in 100 g of water and freezes at -0·24°C. Calculate…
A solution containing $0\cdot5\text{ g}$ of $\text{KCl}$ dissolves in $100\text{ g}$ of water and freezes at $-0\cdot24^\circ\text{C}$. Calculate the degree of dissociation of the salt. ($K_f\text{ for water} = 1\cdot86^\circ\text{C}$).
Answer
Answer
Formulas used: Depression of freezing point: dTf = i*Kf*m with Kf = 1.86 K kg mol^-1, M2 = 74.5 g mol^-1, w1 = 100 g, w2 = 0.5 g, dTf = 0.24 K, so i = 1.923 Degree of dissociation from i: alpha = (i - 1)/(nu - 1) with i = 1.923, nu = 2, so alpha = 0.923
$\Delta T_f = 0\cdot24\ \mathrm{K}$; $M_2(\mathrm{KCl}) = 74\cdot5\ \mathrm{g\ mol^{-1}}$
$\Delta T_f = \dfrac{i K_f w_2 \times 1000}{M_2 w_1}$, so $i = \dfrac{0\cdot24 \times 74\cdot5 \times 100}{1\cdot86 \times 0\cdot5 \times 1000} = 1\cdot92$
KCl gives 2 ions ($\nu = 2$): $i = 1 + \alpha$, so $\alpha = 1\cdot92 - 1 = 0\cdot92$
Final answer: 0.92