CH₄ is adsorbed on 1g charcoal at 0°C following the Freudlich adsorption isotherm. 10.0 mL of CH₄ is adsorbed at 100 mm of Hg, whereas 15.0 mL is adsorbed at 200 mm of Hg. The volume of CH₄ adsorbed at 300 mm of Hg is 10^x mL. The value of x is ______* 10^-2.

(Nearest integer)

[Use log₁₀ 2 = 0.3010, log₁₀3 = 0.4771]

$\left(\frac{x}{m}\right)=K{P}^{1/n}$           

$\frac{10}{1}=K{\left(100\right)}^{1/n}................\left(1\right)$     

$\frac{15}{1}=K{\left(200\right)}^{1/n}................\left(2\right)$

$\frac{V}{1}=K{\left(300\right)}^{1/n}...............\left(3\right)$

Dividing (2) by (1)

$\frac{15}{10}={\left(\frac{2}{1}\right)}^{1/n}$ = (2)^1/n

$\frac{V}{10}=10^{0.279}$

$V=10*10^{0.279}=10^{1.279}=10^x$

$\therefore x=1.279=127.91*10^{-2}\approx 128*10^{-2}$

Ans. = 128