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JEE Main 2015 · 4 Apr · Shift 1, Chemistry question 40

3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is :

Choose one option
Answer and worked solution

Official answer: A

Acid adsorbed: 50×10−3×(0.060−0.042)=9×10−450\times10^{-3}\times(0.060-0.042)=9\times10^{-4} mol, i.e. 9×10−4×60=549\times10^{-4}\times60=54 mg. Per gram of charcoal: 543=18\frac{54}3=18 mg.

Topics:AdsorptionConcentration terms
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