25 cm\(^3\) of 0.02 M KOH neutralized 0.03 g of a monobasic organic acid having the general formula C\(_n\)H\(_{2n + 1}\)COOH. The molecular formula of the acid is?
(C = 12, H = 1, O = 16)
HCOOH
C2H5COOH
CH3COOH
C3H7COOH
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Discussions (17)

if you really understand what you are doing it won't take you any time just less than 2 minutes............you can also use the easiest method
n= CรV/1000
n= 0.02ร25/1000
n=5ร10-โดmoles
n=m/M
M=m/n
M=0.03/0.0005=60g
CnH2n+1COOH=60
12n+2n+1+12+16+16+1=60
14n+46=60
14n=60-46=14
n=14/14=1
CH3COOH---------C.

Jamb wicked!..How many minutes una want make we spend 4dis kind qstns on the "D DAY"?

i think the answer is B cos they used the conc of the base which was 0.78 instead of that of the acid which is 1.2 while finding for the RMM of the acid ... and after correcting and using the right values the answer turns out to be C2H5COOH


