Question

The density of 100 g of gold is 19, 300 kg/m3. If the temperature increases from its freezing point to its boiling point, what will be the new volume of gold (B of gold is three times its a gold = 1.42 x 107/C)​

Answer 1

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Answer 2

Explanation:

Volume of gold present in solution,

=

Densityofgold

Massofgold

=

19g/cm

3

1.9×10

−4

g

=0.1×10

−4

cm

3

For spherical particles of gold with a radius equal to 10nm

The volume of each particle:

3

4

πr

3

=

3

4

×

7

22

×(10×10

−7

cm)

3

=

21

88

×10

−18

cm

3

Number of gold particles present:

=

Volumeofeachparticle

Volumeofgoldinsolution

=

21

88

×10

−18

cm

3

0.1×10

−4

cm

3

=

88

21

×10

13

=2.4×10

12

2.4×10

12

particles of gold are presnt in 1000cm

3

(1 litre).

∴ Number of particles present per mm

3

=

10

6

2.4×10

12

[1L=10

6

mm

3

]

=2.4×10

6