If atomic weights of zinc and oxygen are $65 \mathrm{amu}$ and $16 \mathrm{amu}$ respectively, the molecular weight of zinc oxide in $\mathrm{g} {\mathrm{mol}}^{-}$ will be :

In $\mathrm{ZnO}:$ At. wt. of $\mathrm{Zn}=65 \mathrm{amu} \mathrm{x} \frac{1.66\mathrm{x}{10}^{-24}\mathrm{g}}{1 \mathrm{amu}}\mathrm{x} 6.023 \mathrm{x} {10}^{23}=65 \mathrm{g}$

                                                    $\left(\because 1\mathrm{amu}=1.66\times {10}^{-24}\mathrm{g}\right)$

At. wt. of oxygen, $\mathrm{O}=16\mathrm{amu}\times \frac{1.66\times {10}^{-24}\mathrm{g}}{1\mathrm{amu}}\times 6.023 \mathrm{x} {10}^{23}=16 \mathrm{g}$

$\therefore$  Mol. wt. of $\mathrm{ZnO}=$At. wt. of $\mathrm{Zn} +$ at. wt. of $\mathrm{O}=65+16=81 \mathrm{g} {\mathrm{mol}}^{-}$

Thus the correct answer is $\left(\mathrm{c}\right).$