Manganese forms non-stoichiometric oxides having the general formula ${MnO}_x$. The value of $x$ for the compound that analyzed 64 % Mn. $($ $Mn=55)$

Given, Molar mass of Mn = 55.

$\%$ Of $Mn=64\%$

 Weight of $Mn{O}_x=55+16x$

Percentage of Mn = $\frac{55}{55+16x}\times 100$

$64=\frac{55}{55+16x}\times 100$

64(55 + 16x) = 5500

$16x=\frac{5500}{64}-55$

X = 1.93

Therefore, value of x is 1.93.