If you dissolved 30 mg in 50 mL (0.05 L), you obtained an initial concentration of 0.6 mg/mL = 600 mg/L. If you obtained a removal efficiency of 50%, for example, your final concentration would be 300 mg/L. Applying mass balance for 30 mg or 0.03 (g), you have:

q= (Ci-Cf).V/mass = (600-300) . 0.05 / 0.03 = 500 mg/g

In addition, if you investigated different times, you will calculate q_t, q_e will be only when the final concentrations achieve the equilibrium (same concentrations at different times). In your case, times with 94, 95 and 95 % are probably your adsorption equilibrium.

Note: 30% of absorbance removal is only equal to 30% of I2 adsorption if the calibration curve was linear.

Regards,

You can use the formula to calculate removal % (adsorption %) at any given point of time

R% = (C - Ct)/C × 100

where C (ppm) is the initial concentration of adsorbate and Ct (ppm) is concentration at time t respectively.

Adsorption % can be converted into mg of adsorbate adsorbed on the surface of catalyst. Take for example your last data point 95% adsorption ratio. That means 30x95/100 = 28.5 mg has been adsorbed.

You can refer to this paper for similar adsorption experiments.

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Disregarding the slight change in volume qe ≈ 570/0.6 mg/g and Ce ≈ 30 mg/L.