3 g of activated chasrcoal 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 :

3 g of activated chasrcoal was added to 50 mL of acetic acid solution

1.	3 g of activated chasrcoal 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 :
Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/18-278913" style="font-weight:bold;" target="_blank" title="Click to know more about 18">18</a> mg <br/><a href="https://interviewquestions.tuteehub.com/tag/36-309156" style="font-weight:bold;" target="_blank" title="Click to know more about 36">36</a> mg <br/>42 mg <br/>54 mg </p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Number of moles of acetic acid adsorbed `= (0.06xx(50)/(1000)-0.042xx(50)/(1000))`<br/>`= (0.9)/(1000)` moles. <br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>` Weight of acetic acid adsorbed `= 0.9xx60 mg`<br/>= 54 mg<br/>Hence, the amount of acetic acid adsorbed per g of charcoal `= (54)/(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)` mg = 18 mg<br/>Hence, option (A) is correct.</body></html>