One gm metal M3+ was discharging by the passage of 1.81 *1023 electrons.What is the atomic weight of metal?

For discharging of metal the reaction is :
M³⁺ + 3e^- $\to$ M
Thus 1 mol of the metal is deposited by 3F = 3$\times$96500 C = 289500 C

​Charge on 1.81$\times$10²³ electrons =  1.602$\times$10^-19 $\times$1.81​$\times$10²³ C
                                                   = 2.9$\times$10⁴ C
289500  C deposit = 1 mol of metal 
Thus 2.9$\times$10⁴ C will deposit = $\frac{1}{289500}\times 2.9\times {10}^4\mathrm{moles} \mathrm{metal}$ = 0.100 moles
0.100 moles =  1 g metal
Thus atomic mass of metal = 10 g/mol