Calculate the degree of dissociation of AlCl3

Question

I'm having trouble figuring out where to go with this question.

Calculate the degree of dissociation of $\ce{AlCl3}$ if molar conductivity of the solution is $31.3\ \mathrm{mS\ m^2\ mol^{-1}}$ and constant $\mathrm{K}$ is $\mathrm{108.96\ mS\ m^2\ mol^{-1}}$.

So the equations I think should be used are

for the dissociation constant

$$\alpha =\frac{\Lambda_m }{ \Lambda_m^0}$$

and

$$\Lambda_m = \Lambda_m^0 - K\sqrt{C}$$ where $\mathrm{K}$ is the positive constant for a given electrolyte.

But I don't know how to find C

The only relationship I've found that's related to molar conductance is $$\mathrm{C = k \frac{A}{L}}$$

where this $\mathrm{k}$ is the conductivity

$$\mathrm{k=\frac{1}{\rho}}$$

where $\rho$ is the specific resistance and $\mathrm{\frac{A}{L}}$ is the cell constant. I feel like at this point there's too many unknowns? What should I do?

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Edit

Well, you can understand my frustration when the instructor updated his question last night, with the inclusion of the concentration in it. $\mathrm{C= 0.025}$

Answer 1

So, now that my instructor fixed his question, answering this was fairly straight forward to me. I'll post my answer here anyway.

Solving for the observed molar conductivity $$\begin{align}\Lambda_\mathrm{m} &=\Lambda_\mathrm{m}^0-\mathrm{K \sqrt{C}} \\&=31.3-108.96\sqrt{0.025} \\&= 14.07\end{align}$$

Solving for the degree of dissociation $$\\\begin{align}\alpha &= \frac{\Lambda_m}{\Lambda_m^0}\\\\&=\frac{14.07}{31.3}\\&=0.44\\&=44.95\mathrm{\% } \end{align}$$

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Edit

He just posted the answer key and this was not the correct answer. I used the molar conductivity of the solution as the standard molar conductivity instead of the observed. The answer was

$$\begin{align}\alpha&=\frac{\Lambda_m}{\Lambda_m^0}\\\\&=\frac{\Lambda_m}{\Lambda_m + K\sqrt{C}}\\\\&=\frac{31.3}{31.3+108.96\sqrt{0.025}}\\\\&=0.643\end{align}$$