‹ Back to the paper
Solve the following.
A 0·06 molar CH₃COOH solution offers a resistance of 55 ohms to a conductivity cell at 25°C. If the…
(i)[3]
A $0\cdot06\text{ molar}$ $\text{CH}_3\text{COOH}$ solution offers a resistance of $55\text{ ohms}$ to a conductivity cell at $25^\circ\text{C}$. If the cell constant is $0\cdot45\text{ cm}^{-1}$ and the molar conductance of $\text{CH}_3\text{COOH}$ at infinite dilution is $398\cdot5\text{ ohm}^{-1}\text{cm}^2\text{mol}^{-1}$. Calculate:
(a) Specific conductance
(b) Molar conductance
(c) Degree of dissociation
(ii)[2]
Calculate the number of coulombs of charge required to deposit $24\cdot35\text{ g}$ of aluminium from a solution containing $\text{Al}^{3+}$ ions.
(Atomic weight of $\text{Al} = 27$)
Answer
Answer (i)
Formulas used: Conductivity from resistance: kappa = Gstar/Rcell with Gstar = 0.45 cm^-1, Rcell = 55 ohm, so kappa = 0.00818 S cm^-1 Molar conductivity: Lm = kappa/C with C = 0.06 mol L^-1, kappa = 0.008182 S cm^-1, so Lm = 136.4 S cm^2 mol^-1 Degree of dissociation from molar conductivities: alpha = Lm/Lm0 with Lm = 136.4 S cm^2 mol^-1, Lm0 = 398.5 S cm^2 mol^-1, so alpha = 0.342
(a) Specific conductance: $\kappa = \dfrac{G^*}{R} = \dfrac{0\cdot45\text{ cm}^{-1}}{55\text{ ohm}} = 8\cdot18\times10^{-3}\text{ ohm}^{-1}\text{cm}^{-1}$
(b) Molar conductance: $\Lambda_m = \dfrac{\kappa \times 1000}{C} = \dfrac{8\cdot18\times10^{-3} \times 1000}{0\cdot06} = 136\cdot4\text{ ohm}^{-1}\text{cm}^2\text{mol}^{-1}$
(c) Degree of dissociation: $\alpha = \dfrac{\Lambda_m}{\Lambda_m^\infty} = \dfrac{136\cdot4}{398\cdot5} = 0\cdot342$
Final answer: 0.342
Answer (ii)
Formula used: Faraday's first law: w = M*Q/(n*F) with F = 96500 C mol^-1, M = 27 g mol^-1, n = 3, w = 24.35 g, so Q = 261086 C
Formula: $Q = nF \times \dfrac{w}{M}$ for $\mathrm{Al^{3+} + 3e^- \rightarrow Al}$, so $n = 3$
Substitution:
$Q = 3 \times 96500\text{ C mol}^{-1} \times \dfrac{24\cdot35\text{ g}}{27\text{ g mol}^{-1}} = 2\cdot61\times10^{5}\text{ C}$
Final answer: 261086 C