# In Freundlich adsorption isotherm, slope of AB line is: Option: 1 n with (n, 0.1 to 0.5) Option: 2 $\frac{1}{n}$ with $\left (\frac{1}{n} = 0 to 1 \right )$ Option: 3 $\log \frac{1}{n}$ with (n<1) Option: 4 $\log n \text{with }(n>1)$

We know,

$\\\mathrm{\frac{x}{m}=k\left ( P \right )^{1/n}} \\\\ \mathrm{log\frac{x}{m}=logk+log\left ( P \right )^{1/n}= logk+\frac{1}{n}log\left ( P \right ) } \\$

The equation of a line is

y = mx+c , where m= slope.

After comparing the equation,

Slope m = 1/n.

So, the slope of AB line is 1/n.

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