Answer:
105.86 grams of AuCl3, if the compound has 6.3 x10^23 atoms of Cl.
Step-by-step explanation:
We are given that the compound has 6.3 x10^23 atoms of Cl.
To find how many molecules of AuCl3 are in the given compound, we divide the compound by 3, i.e;
[tex]\frac{6.3 \times 10^{23} }{3}[/tex] = [tex]2.1\times 10^{23}[/tex] molecules of AuCl3.
Now, as we know that 1 mole of AuCI3 has [tex]6.022 \times 10^{23}[/tex] molecules.
So, the moles that our compound has is given by;
= [tex]\frac{2.1 \times 10^{23} }{6.022 \times 10^{23} }[/tex] = [tex]\frac{2.1}{6.022}[/tex] = 0.349 mole AuCI3
Also, the molar mass of AuCI3 = 303.33 g/mole
So, the molar mass of 0.349 moles AuCI3 = [tex]303.33 \times 0.349[/tex]
= 105.86 g
Hence, 105.86 grams of AuCl3, if the compound has 6.3 x10^23 atoms of Cl.
