Respuesta :
Solution :
It is given that :
The formula of the compound = [tex]$XBr_3$[/tex]
And that 4.70 g of the sample contains [tex]$4.834 \times 10^{-2}$[/tex] mol of Br.
It means that :
1 mol of [tex]$XBr_3$[/tex] contains = 3 mol of Br
∴ 3 mol of [tex]$Br^-$[/tex] contain in 1 mol of [tex]$XBr_3$[/tex]
[tex]$4.834 \times 10^{-2}$[/tex] mol of [tex]$Br^-$[/tex] contains [tex]$1.576 \times 10^{-2}$[/tex] mol of [tex]$XBr_3$[/tex]
Thus the mol of [tex]$XBr_3$[/tex] = [tex]$1.576 \times 10^{-2}$[/tex] mol
The given mass is = 4.700 g
Therefore, the [tex]$\text{molar mass}$[/tex] of [tex]$XBr_3$[/tex] [tex]$=\frac{\text{mass}}{\text{mol}}$[/tex]
[tex]$=\frac{4.700 \ g}{1.576 \times 10^{-2} \ mol}$[/tex]
= 298.4 g/mol
So [tex]$\text{molar mass}$[/tex] of [tex]$XBr_3$[/tex] = [tex]$\text{molar mass}$[/tex] of X + 3 x [tex]$\text{molar mass}$[/tex] of Br
298.4 g/mol = [tex]$\text{molar mass}$[/tex] of X + 3 x 79.90 g/mol
298.4 g/mol = [tex]$\text{molar mass}$[/tex] of X + 239.7 g/mol
[tex]$\text{molar mass}$[/tex] of X = 58.71 g/mol (since 1 amu = 1 g/mol)
Therefore the atomic mass of the unknown metal = 58.71 g/mol
So the unknown meta is Nickel.
