25 cm3 of 0 02 M KOH neutralized 0 03 g of a mo... - JAMB Chemistry 1997 Question
25 cm3 of 0.02 M KOH neutralized 0.03 g of a monobasic organic acid having the general formula CnH2n + 1 COOH. The molecular formula of the acid is?
(C = 12, H = 1, O = 16)
(C = 12, H = 1, O = 16)
A
HCOOH
B
C2H5COOH
C
CH3COOH
D
C3H7COOH
correct option: c
Conc. of KOH = 39 x 0.02 = 0.78 g/dm3
25cm3 require 0.03 g of CnH2n + 1 COOH 1000CM3 → x x = (1000) / (25) x 0.03 = 1.2g
CnH2n + 1COOH + KOH → CnH2n + 1 COOK + H2O
from (MAVA) / (MB x VB) = (1) / (1) (MA x 1.2) / (0.02 x 0.78) = (1) / (1)
MA = 0.013
Con. = Molarity x Molecular Mass
Molecular Mass = (0.78) / (0.013) x 60
CnH2n + 1 COOH = 60
12n + 1 (2n + 1) + 12 + 32 + 1 = 6
14n = 60 - 46; n = (14) / (14) = 1
∴ CH3COOH
25cm3 require 0.03 g of CnH2n + 1 COOH 1000CM3 → x x = (1000) / (25) x 0.03 = 1.2g
CnH2n + 1COOH + KOH → CnH2n + 1 COOK + H2O
from (MAVA) / (MB x VB) = (1) / (1) (MA x 1.2) / (0.02 x 0.78) = (1) / (1)
MA = 0.013
Con. = Molarity x Molecular Mass
Molecular Mass = (0.78) / (0.013) x 60
CnH2n + 1 COOH = 60
12n + 1 (2n + 1) + 12 + 32 + 1 = 6
14n = 60 - 46; n = (14) / (14) = 1
∴ CH3COOH
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