JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Surface Chemistry

CH_{4}

is adsorbed on 1 g charcoal at

0^{\circ} C

following the Freundlich adsorption isotherm. 10.0 mL of

CH_{4}

is adsorbed at 100 mm of Hg , whereas 15.0 mL is adsorbed at 200 mm of Hg . The volume of

CH_{4}

adsorbed at 300 mm of Hg is

1 0^{x} mL

. The value of

x

is ____

\times 1 0^{- 2}

.
(Nearest integer)
[Use

lo g_{10} 2 = 0.3010, lo g_{10} 3 = 0.4771

]

Answer: 128

Solution

The Freundlich isotherm

\frac{x}{m} = k P^{1/ n}

gives

lo g V = lo g k + \frac{1}{n} lo g P

. Using the two readings,

lo g 15 - lo g 10 = \frac{1}{n} (lo g 200 - lo g 100)

, so

\frac{1}{n} = \frac{lo g 1.5}{lo g 2} = \frac{0.4771 - 0.3010}{0.3010} = 0.585

. At

300

mm Hg,

lo g V_{3} = lo g 10 + 0.585 lo g 3 = 1 + 0.585 \times 0.4771 = 1.279

, so

V_{3} = 1 0^{1.279}

mL. With the volume written as

1 0^{x}

mL,

x = 1.279

, and

x \times 1 0^{2} = 127.9

, which rounds to

128

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Integer answer