The initial mass of acetic acid is $$0.45\ \text{g}$$. With molar mass $$60\ \text{g mol}^{-1}$$, the initial number of moles is
$$n_0=\frac{0.45}{60}=0.0075\ \text{mol}$$
The solution volume is $$50\ \text{mL}=0.05\ \text{L}$$, so the initial concentration is
$$C_0=\frac{n_0}{0.05}=0.15\ \text{mol L}^{-1}$$
After the mixture is shaken with $$1.0\ \text{g}$$ of charcoal, the pH of the equilibrium solution is 3.0, i.e.
$$[H^+]=10^{-3}\ \text{mol L}^{-1}$$
For a weak monoprotic acid $$\mathrm{HA}$$ of equilibrium concentration $$C$$,
$$K_a=\frac{[H^+]^2}{C-[H^+]} \quad -(1)$$
Using $$K_a = 1.0\times10^{-5}$$ and $$[H^+]=10^{-3}$$, equation $$-(1)$$ gives
$$1.0\times10^{-5}=\frac{(10^{-3})^2}{C-10^{-3}}$$
$$\Rightarrow\;1.0\times10^{-5}(C-10^{-3})=10^{-6}$$
$$\Rightarrow\;C-10^{-3}=0.10$$
$$\Rightarrow\;C\approx0.101\ \text{mol L}^{-1}\;(\text{very close to }0.10\ \text{mol L}^{-1})$$
Moles of acetic acid that remain in solution:
$$n_{\text{eq}} = C \times 0.05 \approx 0.005\ \text{mol}$$
Moles adsorbed on charcoal:
$$n_{\text{ads}} = n_0 - n_{\text{eq}} = 0.0075-0.005 = 0.0025\ \text{mol}$$
Mass adsorbed:
$$x = n_{\text{ads}}\times60 = 0.0025\times60 = 0.15\ \text{g}$$
Given $$m = 1.0\ \text{g}$$ of charcoal,
$$\frac{x}{m} = \frac{0.15}{1.0}=0.15$$
The Freundlich isotherm is $$\dfrac{x}{m}=kC^{1/n}$$. The slope of the plot of $$\log_{10}(x/m)$$ versus $$\log_{10}C$$ is the exponent $$1/n$$. Because the slope is given to be 1,
$$\frac{1}{n}=1\;\Rightarrow\;n=1$$
Hence the equation reduces to
$$\frac{x}{m}=kC$$
Therefore,
$$k=\frac{x/m}{C}=\frac{0.15}{0.10}=1.50\ \text{L mol}^{-1}$$
Final answer: $$1.50$$
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