A dye solution (1.00 L, 100 mg L^-1) is shaken with 2.0 g of an adsorbent. At equilibrium, x/m=ksqrtC with k=5.0 mg g^-1 (mg L^-1)^-1/2, where C is the equilibrium concentration in mg L^-1. Find C (in mg L^-1, one decimal).

JEE Advanced levelNumerical answerDifficulty 3 of 3sc046

A dye solution ( $1.00 L$ , $100 mg L^{- 1}$ ) is shaken with $2.0 g$ of an adsorbent. At equilibrium, $x / m = k C$ with $k = 5.0 mg g^{- 1} (mg L^{- 1})^{- 1/2}$ , where $C$ is the equilibrium concentration in $mg L^{- 1}$ . Find $C$ (in $mg L^{- 1}$ , one decimal).

Numerical answer type, answer in mg/L, tolerance 0.2.

Mass balance:

(100 - C) \times 1.00 = 2.0 \times 5.0 C

. Let

s = C

s^{2} + 10 s - 100 = 0

, so

s = (- 10 + 500) /2 = 6.1803

. Hence

C = s^{2} = 38.2 mg L^{- 1}