The graph between $l n (\frac{x}{m})$ and $l n (p)$ is a straight line inclined at an angle of $45^{\circ}$ where x is the mass of the solute absorbed and m is the mass of the absorbent while p is the pressure of the gas in atm. When the pressure is 0.5 atm and $l n (k) = 0.693$ where k is the Freundlich constant, then what amount of solute will be adsorbed by 12 g of the adsorbent?

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Solution

$l n (\frac{x}{m}$ ) = $l n (k)$ + $\frac{1}{n} l n (p)$

$l n (\frac{x}{m}$ )= 0.693 + $l n (0.5)$ =0

$\frac{x}{m}$ =1

Therefore, 12 g of the adsorbent, adsorbs12 g of the solute.

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ln(xm) = ln(k) + 1nln(p) ln(xm)= 0.693 + ln(0.5)=0 xm=1 Therefore, 12 g of the adsorbent, adsorbs12 g of the solute.

$l n (\frac{x}{m}$ ) = $l n (k)$ + $\frac{1}{n} l n (p)$

$l n (\frac{x}{m}$ )= 0.693 + $l n (0.5)$ =0

$\frac{x}{m}$ =1

Therefore, 12 g of the adsorbent, adsorbs12 g of the solute.