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JEE Advanced levelSingle correct optionDifficulty 3 of 3sc041

A dye in $0.50 L$ of solution is adsorbed on $2.0 g$ of adsorbent obeying $x / m = k C$ (Freundlich with $1/ n = 1$ , $C$ = equilibrium concentration) with $k = 0.15 L g^{- 1}$ . The fraction of the dye removed from solution is:

A $0.625$
B $0.375$
C $0.600$
D $0.300$

Answer: B

Mass balance:

(C_{0} - C) V = mk C

, so

C_{0} / C = 1 + mk / V = 1 + (2.0 \times 0.15) /0.50 = 1.60

. Fraction removed

= 1 - 1/1.60 = 0.375

. The value

0.625

is the fraction remaining.

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