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50 ml of 1M oxalic acid is shaken with 0.5 g of wood charcoal. The final concentration of solution after adsorption is 0.5 M .Calculate the amount of oxalic acid adsorbed per gram of charcoal?
Total moles of oxalic acid = 50 mmolFinal concentration = 0.5 MFinal no. of moles = 0.5/50 ml = 10 mmolamount of oxalic acid adsorbed per gram = 40 mmol/0.5 = 80 mmol/g = 80 X 126 / 1000 = 10.08 gram of oxalic acid / gram of charcoalThanksBharat BajajIIT Delhiaskiitians faculty
Initial concentration =1MFinal concentration = 0.5 MVolume = 0.05 LInitial no of moles = 1/20Initial mass of oxalic acid = 6.3 gFinal mass of oxalic acid = 3.15gAmount of oxalic acid adsorbed per gram of charcoal =3.15/0.5 =6.3gVinu Thanks
1M oxalic acid solutions mean; 1mole of oxalic acid presents in 1000mL solution 50mL of 1M solutions will contain oxalic acid =126/1000×50=6.6g
Initial amount of oxalic acid present in 50 ml=126/1000×50=6.3g.Amount of oxalic acid in 50ml after adsorption=3.15gAmount of oxalic acid adsorbed per gram of charcoal =3.15×2=6.3g
Given, Volume of Oxalic Acid = 50ml = 0.05LInitial Concentration- 1MFinal Concentration- 0.5MSo, Initial Moles- (0.05*1)= 0.05molesFinal Moles- (0.05*0.5)= 0.025molesChange in Moles= 0.05-0.025 = 0.025moles Now, Amount adsorbed=So, (0.025*90.03)/0.5 = 4.5015 gramsThanks,Sanskar Goyal
Dear student,Please find the attached solution to the problem. Initial moles of oxalic acid = 1M * 0.05L = 0.05 molesFinal concentration = 0.5M * 0.05L = 0.025 molesMoles of oxalic acid adsorbed = 0.05 – 0.025 = 0.025 molesMolar mass of oxalic acid = 90gMass of oxalic acid adsorbed = 0.025 * 90 = 2.25gHence oxalic acid adsorbed per gram of charcoal = 2.25/0.5 = 4.5 Hope it helps.Thanks and regards,Kushagra
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