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JEE Main levelSingle correct optionDifficulty 2 of 3ck123

Acid-catalysed hydrolysis of an ester is followed by titrating 5 mL aliquots with NaOH. The titres are 12.0 mL at t = 0, 18.0 mL at t = 30 min and 36.0 mL at infinite time. The first-order rate constant in $mi n^{- 1}$ is

A $0.0096$
B $0.0135$
C $0.0366$
D $0.0231$

Answer: A

V_{\infty} - V_{t}

is proportional to the ester left.

k = \frac{1}{t} ln \frac{V _{\infty} - V _{0}}{V _{\infty} - V _{t}} = \frac{1}{30} ln \frac{24.0}{18.0} = 0.0096 mi n^{- 1}

.

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