The hydrolyiss of methyl acetate in aqueous solution is has been studied by titrating the liberated acetic acid against `NaOH`. The concentration of ester at different times is given below:
`|{:("t (min)",0,30,60,90),(C (Mol L^(-1)),0.8500,0.8004,0.7538,0.7096):}|`
Show that it follows a pseudo first order reaction, as the concentration of water remains nearly constant `(55 mol L^(-1))` during the course of the reaction . What is the value of `k'` in the equation ?
Rate `= k'[CH_(3)COOCH_(3)][H_(2)O]`

Explanation

To demonstrate that the hydrolysis of methyl acetate follows a pseudo first-order reaction and to calculate the value of \( k' \), we will follow these steps:

### Step 1: Understanding the Reaction
The hydrolysis of methyl acetate (\( CH_3COOCH_3 \)) in aqueous solution produces acetic acid (\( CH_3COOH \)) and methanol (\( CH_3OH \)). The reaction can be simplified as:
\[ CH_3COOCH_3 + H_2O \rightarrow CH_3COOH + CH_3OH \]

### Step 2: Rate Law Expression
The rate of the reaction can be expressed as:
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