In a first order reaction, t_1/2 = 245 days of compound A. After x days, 75% of A remains.

Calculate the value of x.

Given: (log 2 = 0.3 ,log 3 = 0.48)

Let us solve the problem step-by-step.

First, let's understand the kinetics of a first-order reaction. The rate constant $k$ is related to the half-life $t_{1/2}$ by the formula:
$k=\frac{\ln \left(2\right)}{t_{1/2}}=\frac{2.303\times \log \left(2\right)}{t_{1/2}}$
Given that the half-life $t_{1/2}=245\text{days}$ and $\log \left(2\right)=0.3$, we have:
$k=\frac{2.303\times 0.3}{245}{\text{day}}^{-1}$

For a first-order reaction, the relation between the initial concentration ${\left[A\right]}_0$ and the concentration remaining at time $t$, ${\left[A\right]}_t$, is:
$t=\frac{2.303}{k}\log \left(\frac{{\left[A\right]}_0}{{\left[A\right]}_t}\right)$

Here, $t=x$ days, and 75% of compound A remains. This means:
${\left[A\right]}_t=75\%\text{of}{\left[A\right]}_0=0.75{\left[A\right]}_0=\frac{3}{4}{\left[A\right]}_0$

Substitute this into the expression for $x$:
$x=\frac{2.303}{k}\log \left(\frac{{\left[A\right]}_0}{\frac{3}{4}{\left[A\right]}_0}\right)=\frac{2.303}{k}\log \left(\frac{4}{3}\right)$

Using the properties of logarithms:
$\log \left(\frac{4}{3}\right)=\log \left(4\right)-\log \left(3\right)=2\log \left(2\right)-\log \left(3\right)$
Using the given values $\log \left(2\right)=0.3$ and $\log \left(3\right)=0.48$:
$\log \left(\frac{4}{3}\right)=2\left(0.3\right)-0.48=0.6-0.48=0.12$

Now, substitute the values of $k$ and $\log (4/3)$ back into the equation for $x$:
$x=2.303\times \left(\frac{245}{2.303\times 0.3}\right)\times 0.12$
We can cancel out $2.303$ from the numerator and denominator:
$x=\frac{245}{0.3}\times 0.12$
Simplifying the expression:
$x=245\times \frac{0.12}{0.3}=245\times 0.4=98$

Therefore, the value of $x$ is 98 days.

The correct option/answer is 98.