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**Hint:**The half lifetime for 4 half-lives will be equal to 4 times the half-life of 1 half-life i.e. 1 day and substituting ${t_{\dfrac{1}{2}}}^\prime = 4(1day) = 4days$ in integrated rate equation for the first-order reaction, we can find the value of $x$ but first of all, we need to find the value of the rate constant $k$. For a first-order reaction, $k = \dfrac{{2.303}}{t}{\log _{10}}\dfrac{a}{{a - x}}$

Here, $a$ is the initial concentration

$x$ is the amount of reactant reacting in time t

**Complete step by step answer:**

Now, we know that the half-life formula for the first-order reaction is,

${t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{k}$

Here, $k$ is rate constant

For 1 half-life,

${t_{\dfrac{1}{2}}} = 1day$ (Given)

$\therefore $For 4 half-lives,

${t_{\dfrac{1}{2}}}^\prime = 4\left( {{t_{\dfrac{1}{2}}}} \right)$

$ = 4(1)$

${t_{\dfrac{1}{2}}}^\prime = 4days$ ……. (Equation number 1)

Now,

${t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{k}$

So, $k = \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}$

$ \Rightarrow k = \dfrac{{0.693}}{1}$

$k = 0.693da{y^{ - 1}}$ ….. (Equation number 2)

For $t = {t_{\dfrac{1}{2}}}$,

${t_{\dfrac{1}{2}}} = \dfrac{{2.303}}{k}{\log _{10}}\dfrac{a}{{a - x}}$

And for $t = {t_{\dfrac{1}{2}}}^\prime $,

${t_{\dfrac{1}{2}}}^\prime = \dfrac{{2.303}}{k}{\log _{10}}\dfrac{{100}}{{100 - x}}$ …… (Equation number 3)

As the reaction completes $x\% $ after 4 half-lives. (Given)

But ${t_{\dfrac{1}{2}}}^\prime = 4days$ …..(From Equation number 1)

And $k = 0.693da{y^{ - 1}}$ …..(From Equation number 2)

Substituting these values in Equation number 3,

$4 = \dfrac{{2.303}}{{0.693}}{\log _{10}}\dfrac{{100}}{{100 - x}}$

$ \Rightarrow 4 = 3.3232\left[ {{{\log }_{10}}\left( {100} \right) - {{\log }_{10}}\left( {100 - x} \right)} \right]$

$ \Rightarrow 4 = 3.3232\left[ {2 - {{\log }_{10}}\left( {100 - x} \right)} \right]$

$ \Rightarrow \dfrac{4}{{3.3232}} = 2 - {\log _{10}}\left( {100 - x} \right)$

$ \Rightarrow 1.2036 = 2 - {\log _{10}}\left( {100 - x} \right)$

$ \Rightarrow 1.2036 - 2 = - {\log _{10}}\left( {100 - x} \right)$

$ \Rightarrow - 0.7964 = - {\log _{10}}\left( {100 - x} \right)$

$ \Rightarrow 0.7964 = {\log _{10}}\left( {100 - x} \right)$

Now, converting logarithmic form to exponential form,

${10^{0.7964}} = 100 - x$

Taking log on both the sides,

$\log \left( {{{10}^{0.7964}}} \right) = \log \left( {100 - x} \right)$

$0.7964 = {\log _{10}}\left( {100 - x} \right)$

Now taking antilog on both sides,

$A.L.\left( {0.7964} \right) = A.L.\left[ {{{\log }_{10}}\left( {100 - x} \right)} \right]$

$6.2574 = 100 - x$

$ \Rightarrow 6.2574 - 100 = x$

$ \Rightarrow - 93.75 = - x$

$x = 93.75$

Hence, the value of $x$ is $93.75$

**So, the correct answer is Option C.**

**Additional Information:**

The concept of half-life was discovered by Ernest Rutherford in 1907. It relates to the time required by radioactive substances for disintegration or the formation of a new substance. It is also known as the half-life period.

**Note:**The rate constant of a second-order reaction is $k = \dfrac{1}{{2at}}\log \dfrac{x}{{a - x}}$. The first-order rate constant depends on the concentration of only 1 among the reactants. We could see that the half-life of the first order is independent of the reactant concentration. The unit of the first-order reaction is ${\sec ^{ - 1}}$. The unit of half-life is seconds.

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