At \(T\)(K), adsorption of gas \((A)\) on solid adsorbent \((B)\) follows Freundlich adsorption isotherm. On \(10\) g of \(B\), adsorption of gas \((A)\) gave the isotherm shown below. What is \(x\) (the quantity of \(A\) adsorbed in one gram of \(B\)) when the pressure of \(A\) is \(1.259\) atm?
\[
(\log 1.259=0.1,\qquad \text{Antilog}(0.1)=1.259)
\]
Show Hint
Freundlich adsorption isotherm is
\[
\boxed{
\frac{x}{m}=kP^{1/n}
}
\]
or
\[
\boxed{
\log\left(\frac{x}{m}\right)=\log k+\frac1n\log P.
}
\]
The slope of the graph gives
\[
\boxed{\frac1n}.
\]
