Answer: There are [tex]17.6\times 10^{23}[/tex] alumnium ions in 1.46 mol [tex]Al_2S_3[/tex]
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
1 mole of [tex]Al_2S_3[/tex] contains = 2 moles of [tex]Al^{3+}[/tex] ions
Thus 1.46 mole of [tex]Al_2S_3[/tex] contains = [tex]\frac{2}{1}\times 1.46=2.92[/tex] moles of [tex]Al^{3+}[/tex] ions
Now 1 mole of [tex]Al^{3+}[/tex] contains = [tex]6.023\times 10^{23}[/tex] [tex]Al^{3+}[/tex] ions
Thus 2.92 moles of [tex]Al^{3+}[/tex] contains = [tex]\frac{6.023\times 10^{23}}{1}\times 2.92=17.6\times 10^{23}[/tex] [tex]Al^{3+}[/tex] ions.
There are [tex]17.6\times 10^{23}[/tex] alumnium ions in 1.46 mol [tex]Al_2S_3[/tex]
