This question is about gold and compounds of gold - AQA - GCSE Chemistry Combined Science - Question 6 - 2020 - Paper 1

To find the diameter of one gold atom, we divide the thickness of the gold foil by the number of atoms:

Diameter = ( \frac{4.00 \times 10^{-7}}{2400} ) m

Calculating this gives:

Diameter = ( 1.6666 \times 10^{-10} ) m, which rounds to ( 1.67 \times 10^{-10} ) m.

From the balanced equation:

2 Au + 3 Cl₂ → 2 AuCl₃

We know:

• 0.175 g of gold will react with ( \frac{3}{2} ) moles of Cl₂.

First, we calculate moles of gold:

Moles of Au = ( \frac{0.175}{197} = 0.000888 ) moles.

Now, moles of Cl₂ needed:

Moles of Cl₂ = ( 0.000888 \times \frac{3}{2} = 0.00133 ) moles.

Now calculating mass of Cl₂:

Mass of Cl₂ = Moles × Molar mass = ( 0.00133 \times 71 = 0.0946 ) g.

Thus, the mass of chlorine needed is 94.6 mg.