Elemental mercury is a silver liquid at room temperature. Its normal freezing point is –38.9 °C, and its molar enthalpy of fusion is ∆Hfusion = 2.29 kJ/mol. What is the entropy change of the system (in J/K) when 5.590 g of Hg(l) freezes at the normal freezing point?

ΔS = -0.272 J/K

Explanation:

Step 1: Data given

Its normal freezing point is –38.9 °C

molar enthalpy of fusion is ∆Hfusion = 2.29 kJ/mol

Mass of Hg = 5.590 grams

Step 2:

ΔG = ΔH - TΔS

At the normal freezing point, or any phase change in general,  

ΔG =0

0  =  Δ

Hfus

−

Tfus  Δ

S

fus

Δ

S

fus = Δ

Hfus

/Tfus

Δ

S

fus = 2290 J/mol / 234.25 K

Δ

S

fus = 9.776 J/mol*K

Since fusion is  from solid to liquid. Freezing is the opposite process, so the entropy change of freezing is -9.776 J/mol*K

Step 3: Calculate moles Hg

Moles Hg = 5.590 grams / 200.59 g/mol

Moles Hg = 0.02787 moles

Step 4: Calculate the entropy change of the system:

Δ

S = -9.776 J/mol*K * 0.02787 moles

ΔS = -0.272 J/K

any far is the region can control the climate

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3