For below graph is $CO$ chemisorption on $Ni$ for a $C{e}_{0\mathrm{.}9}T{i}_{0\mathrm{.}1}{O}_2$ supported $Ni$ catalysts, where the $y-$intercept for total adsorbed amount is $53\mathrm{.}259mo\frac{l}{g},$ and the $y-$intercept for reversible adsorbed amount is $28\mathrm{.}896mo\frac{l}{g}.$ We know there is a total of $0\mathrm{.}16g$ of sample for measurement, $5\%$ of the weight of catalysts is Ni$.$ Calculate the dispersion. $($Ni MW: $58\mathrm{.}69).$ And what is the particle diameter? $(1$ point$)$

Dispersion $=$ exposed atoms $($chemisorbed$)\frac{?}{?}$ total atoms

Chemisorbed molecule $=V_{\text{t}\text{o}\text{t}\text{a}\text{l}\text{m}\text{o}\text{n}\text{o}\text{l}\text{a}\text{y}\text{e}\text{r}\text{a}\text{d}\text{s}\text{o}\text{r}\text{p}\text{t}\text{i}\text{o}\text{n}}-V_{\text{r}\text{e}\text{v}\text{e}\text{r}\text{s}\text{i}\text{b}\text{l}\text{e}(physiosorbed)\text{m}\text{o}\text{n}\text{o}\text{l}\text{a}\text{y}\text{e}\text{r}\text{a}\text{d}\text{s}\text{o}\text{r}\text{p}\text{t}\text{i}\text{o}\text{n}}$

Particle diameter $=\frac{1}{?}$ dispersion