JEE Advanced levelOne or more correct optionsDifficulty 3 of 3sc054
0.50L of a dye solution of 80mgL−1 is stirred with 2.0g of adsorbent. The equilibrium concentration is 32mgL−1. Assume x/m=kC (Freundlich, 1/n=1). Which statements are correct?
Ax/m=12mgg−1
B60% of the dye is removed
Ck=0.150Lg−1
DWith 4.0g of adsorbent in the same volume and initial concentration, the equilibrium concentration would be 20mgL−1
Answer: A, B, D
Dye removed =(80−32)×0.50=24mg, so x/m=24/2.0=12mgg−1 and the fraction removed is 48/80=60%. k=12/32=0.375Lg−1, so statement 3 is wrong. With 4.0g: C=80/(1+4.0×0.375/0.50)=80/4=20mgL−1.