Question

Question 3 5 pts Gold nanoparticles are typically prepared by the reduction of aurochloric acid (HAUCla) with boiling sodium citrate. The sodium citrate is used in excess and it is assumed that 100% of the Au atoms end up as part of the nanoparticles. If you start with a 1.00% (w/w) aqueous solution of HAC14:3H20 (aurochloric acid is typically available as the trihydrate) and the average particle diameter is 20.0 nm, what is the concentration of Au nanoparticles in the final solution? Express your answer in both number of particles per 100.0 mL and in molarity. Hint: you will need to use the density of Au and the volume of a sphere to solve this problem. Your answer will be in scientific notation. The answer box isn't very smart. A number such as 2.05 x 1012 needs to be entered as 2.05 x 10(12). Concentration equals particles per 100 ml Concentration equals

Answer 1

Let's start with 100 grams of HAuCl₄.3H₂O solution.

Since the solution is 1% w/v, we can say that per 100 mL of the solution, we get 1 grams of HAuCl₄.3H₂O as solute.

Molecular weight of HAuCl₄.3H₂O is 394 and weight of Au in it is 197.

So, 394 grams of HAuCl₄.3H₂O contains 197 g of Au

Hence, 1 gram of HAuCl₄.3H₂O contains (197/394) grams or 0.5 g of Au.

Now, assuming the gold atom to be spherical with 20.0 nm diameter or 10.0 nm or 10^-6 cm radius.

The volume of the gold atom be thus calculated as=

(radius)

- (10=6)3

= 4.187 X 10-18 cm

Density of Au is 19.3 gcm^-3.

As we know,

Mass of a gold atom = Density of gold * Volume of a gold atom

grams

= 4.187 x 10-18 x 19.3 = 8.08 x 10-17

Number of gold atoms contained in 0.5 g Au

= 0.5 = (8.08 x 10-17)

= 6.19 x 1015

Thus
6.19 x 1015
atoms of Au are present in 100 mL of the solution.

In terms of molarity, we need to count the number of moles of Au in 1L i.e. 1000 mL of the solution.

100 mL of solution has
6.19 x 1015
atoms or
6.19 x 1015 6.023 x 1023
moles or
1.03 x 10-8
moles of Au.

Therefore ,

1000 mL of solution has moles of Au.

10 x 1.03 x 10-8 = 1.03 x 10-7

Hence, molarity = number of moles / volume of solution in L =
1.03 x 10-7
M