Question

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How many grams of AuCl3, if the compound has 6.3 x10^23 atoms of Cl?

Answer 1

105.86 grams of AuCl3, if the compound has 6.3 x10^23 atoms of Cl.

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;

`(6.3 * 10^(23) )/(3)` =
`2.1* 10^(23)` molecules of AuCl3.

Now, as we know that 1 mole of AuCI3 has
`6.022 * 10^(23)` molecules.

So, the moles that our compound has is given by;

=
`(2.1 * 10^(23) )/(6.022 * 10^(23) )` =
`(2.1)/(6.022)` = 0.349 mole AuCI3

Also, the molar mass of AuCI3 = 303.33 g/mole

So, the molar mass of 0.349 moles AuCI3 =
`303.33 * 0.349`

= 105.86 g

Hence, 105.86 grams of AuCl3, if the compound has 6.3 x10^23 atoms of Cl.