What is the molar mass of 0.500 g of unknown monos optic acid if 22.5 mL of 0.15 M NaOH solution was required to neutralize it?
回答 (2)
2016-06-20 6:47 pm
Let HA be the formula of the acid.
HA + NaOH → NaA + H₂O
OR: Mole ratio HA : NaOH = 1 : 1
No. of moles of NaOH = (0.15 mol/L) × (22.5/1000 L) = 0.003375 mol
No. of moles of HA = 0.003375 mol
Molar mass of HA = (0.500 g) / (0.003375 mol) = 148 g/mol
HA + NaOH → NaA + H₂O
OR: Mole ratio HA : NaOH = 1 : 1
No. of moles of NaOH = (0.15 mol/L) × (22.5/1000 L) = 0.003375 mol
No. of moles of HA = 0.003375 mol
Molar mass of HA = (0.500 g) / (0.003375 mol) = 148 g/mol
2016-06-20 6:46 pm
(0.0225 L) x (0.15 mol/L NaOH) x (1 mol acid/ 1 mol NaOH) = 0.003375 mol acid
(0.500 g) / (0.003375 mol) = 148 g/mol
(0.500 g) / (0.003375 mol) = 148 g/mol
收錄日期: 2021-04-18 15:08:46
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160620103817AA5VRw4