Concentration of KOH = 39 x 0.02 = 0.78 g/dm\(^3\)

25cm\(^3\) required 0.03 g of C\(_n\)H\(_{2n + 1}\)COOH

1000cm\(^3\) will require  X

 \(\frac{{1000}\times{0.03}}{25}\) = 1.2g

C\(_n\)H\(_{2n + 1}\)COOH + KOH → C\(_n\)H\(_{2n + 1}\)COOK + H\(_2\)O

Recall that \(\frac{M_A V_A}{M_B V_B}\) = \(\frac{1}{1}\)

\(\frac{{M_A}\times{1.2}}{{0.02}\times{0.78}}\) = \(\frac{1}{1}\)

M\(_A\) = 0.013
Con. = Molarity x Molecular Mass

Molecular Mass = (0.78) / (0.013) x 60

C\(_n\)H\(_{2n + 1}\)COOH = 60

12n + 1 (2n + 1) + 12 + 32 + 1 = 60

14n = 60 - 46

n =\(\frac{14}{14}\) = 1

Substituting n = 1 into C\(_n\)H\(_{2n + 1}\)COOH

We'll have CH\(_3\)COOH

∴ CH\(_3\)COOH