Question 11

Freundlich isotherm: $\frac{x}{m} = k C^{1/n}$. Let $y = \frac{x}{m}$.

Case 1: $4 = k(10)^{1/n} \Rightarrow \log 4 = \log k + \frac{1}{n} \log 10 \Rightarrow 0.6 = \log k + \frac{1}{n}$ ... (i)

Case 2: $10 = k(16)^{1/n} \Rightarrow \log 10 = \log k + \frac{1}{n} \log 16 \Rightarrow 1 = \log k + \frac{4 \times 0.3}{n} = \log k + \frac{1.2}{n}$ ... (ii)

Subtracting (i) from (ii):

$1 - 0.6 = (\log k + \frac{1.2}{n}) - (\log k + \frac{1}{n})$

$0.4 = \frac{0.2}{n} \Rightarrow n = 0.5$, so $\frac{1}{n} = 2$.

From (i): $0.6 = \log k + 2 \Rightarrow \log k = -1.4 \Rightarrow k = 10^{-1.4}$.

For $C = 20 \text{ mg g}^{-1}$: $y = k(20)^{1/n} = 10^{-1.4} \times (20)^2 = 10^{-1.4} \times 400$.

Alternatively, using the ratio: $\frac{y}{4} = \left(\frac{20}{10} \right)^{1/n} = 2^2 = 4$.

$y = 4 \times 4 = 16 \text{ mg g}^{-1}$.