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Choose the Best Answer · Q1

Q.For the Freundlich adsorption isotherm, a graph of log⁡(x/m)\log(x/m) is plotted against log⁡p\log p. The slope of the line and its y-axis intercept respectively correspond to

(a) 1/n1/n, kk
(b) log⁡(1/n)\log(1/n), kk
(c) 1/n1/n, log⁡k\log k
(d) log⁡(1/n)\log(1/n), log⁡k\log k
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✓ Free question

Step 1. The Freundlich isotherm is xm=kp1/n\dfrac{x}{m} = kp^{1/n}, which is not linear in p, so it is tested by taking logarithms of both sides.

Step 2. Taking logs: log⁡xm=log⁡k+1nlog⁡p\log\dfrac{x}{m} = \log k + \dfrac{1}{n}\log p. Comparing this with the straight-line form y=mx+cy = mx + c (with y=log⁡(x/m)y = \log(x/m) and x=log⁡px = \log p), the coefficient of log⁡p\log p is the slope, and the constant term is the y-intercept.

Step 3. So the slope of the log⁡(x/m)\log(x/m) vs log⁡p\log p line is 1/n1/n, and the y-intercept is log⁡k\log k — not kk itself, since the equation was logarithmic throughout.

✓Final answer

Slope =1/n= 1/n and intercept =log⁡k= \log k, option (c).

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