If 1/6 in place of 1/12 of mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will 

(3) Avogadro's number depends upon the basis of the atomic weight scale.

$\begin{array}{l}\because \mathrm{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}12\mathrm{\hspace{0.17em}}\mathrm{g}\mathrm{\hspace{0.17em}}\mathrm{C}=6.023\mathrm{\hspace{0.17em}}\times \mathrm{\hspace{0.17em}}{10}^{23}\mathrm{\hspace{0.17em}}\mathrm{atoms}\\ \therefore \mathrm{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}6\mathrm{\hspace{0.17em}}\mathrm{g}\mathrm{\hspace{0.17em}}\mathrm{C}=\frac{6.023\mathrm{\hspace{0.17em}}\times \mathrm{\hspace{0.17em}}{10}^{23}\times 6\mathrm{\hspace{0.17em}\hspace{0.17em}}}{12}\\ =\frac{1}{2}\mathrm{\hspace{0.17em}}\times \mathrm{\hspace{0.17em}}6.023\mathrm{\hspace{0.17em}}\times {10}^{23}\mathrm{\hspace{0.17em}\hspace{0.17em}}\mathrm{atoms}=\mathrm{Av}.\mathrm{No}.\\ \mathrm{Or}\mathrm{\hspace{0.17em}\hspace{0.17em}}1\mathrm{\hspace{0.17em}\hspace{0.17em}}\mathrm{amu}=\frac{1}{\mathrm{N}}=\frac{2}{6.023\mathrm{\hspace{0.17em}\hspace{0.17em}}\times \mathrm{\hspace{0.17em}}{10}^{23}}\mathrm{\hspace{0.17em}}=\mathrm{\hspace{0.17em}}3.3\mathrm{\hspace{0.17em}}\times {10}^{-24}\mathrm{g}\end{array}$

Let molar mass of an element be M amu

Then 

 $\begin{array}{l}\mathrm{M} \mathrm{amu} = \mathrm{M}\times 3.3\times {10}^{-24}\text{\hspace{0.17em}\hspace{0.17em} g}\\ \therefore \mathrm{Mass}\mathrm{of}\mathrm{mole}=\mathrm{M}\times 3.3\times {10}^{-24}\times \mathrm{Av}.\mathrm{No}.\\ =\mathrm{M}\times 3.3\times {10}^{-24}\times \frac{1}{2}\times 6.023\times {10}^{23}=\mathrm{Mg}\end{array}$