500ml of a solution contain 12.6 grams of oxalic acid (mol. at.=126) 10ml of this solution is diluted to woo no in a flask what is the molarity of the resultant solution
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Total moles of oxalic acid = 50 mmol
Final concentration = 0.5 M
Final no. of moles = 0.5/50 ml = 10
mmol
amount 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 charcoal
Thanks
Total moles of oxalic acid = 50 mmol
Final concentration = 0.5 M
Final no. of moles = 0.5/50 ml = 10
mmol
amount 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 charcoal
Thanks
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