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The specific conductivity of 0·0025 M acetic acid solution is 7·95 × 10⁻⁵ ohm⁻¹ cm⁻¹. Calculate the…
The specific conductivity of $0\cdot0025\text{ M}$ acetic acid solution is $7\cdot95 \times 10^{-5}\text{ ohm}^{-1}\text{ cm}^{-1}$. Calculate the molar conductivity of this solution. If molar conductance for acetic acid is $389\cdot5\text{ ohm}^{-1}\text{ cm}^2\text{ mol}^{-1}$ at infinite dilution ($\Lambda_m^\circ$), what would be its degree of dissociation?
Answer
Answer
$\Lambda_m=\frac{\kappa\times1000}{C}=\frac{7.95\times10^{-5}\times1000}{0.0025}=31.8\text{ ohm}^{-1}\text{ cm}^2\text{ mol}^{-1}$; $\alpha=\frac{\Lambda_m}{\Lambda_m^\circ}=\frac{31.8}{389.5}=0.0816$
Final answer: 31.8 ohm^-1 cm^2 mol^-1