A pure substance with a molar mass of 0.122 kg/mol has solid and liquid phases w

Question

A pure substance with a molar mass of 0.122 kg/mol has solid and liquid phases with

densities of 1.08Å~103 kg/m3 and 1.01Å~103 kg/m3, respectively, at its standard melting point of

427. K. The standard boiling point for this pure substance is 550. K, with a molar enthalpy

of vaporization of 2.20Å~104 J/mol.

a. The melting point for this pure substance increases to 429. K if the pressure is increased

to 1.20Å~107 Pa. Determine fusSm (in J/mol-K) for the substance, assuming both fusSm & fusVm are independent of temperature and pressure.

b. Determine the boiling point (in K) for this pure substance when the pressure is 2.00 bar.

c. The standard boiling point rises by 2.00 K when 2.00 mol of naphthalene is dissolved in 10.0 kg of the substance. Determine whether naphthalene experiences net dissociation, net association, or neither in this solvent based on the apparent van’t Hoff factor for naphthalene in this substance, assuming the solution is sufficiently dilute.

Explanation / Answer

a. We will use Clausius Clapeyron equation to find enthalpy of fusion

for solid <---> liquid equilibrium that is melting , Clausius-Clapeyron equation is written as

P2 - P1 = delta H (fus) / delta V * ( ln T2 - ln T1)

we have P1 = 1 atm which is 1.013 x 10^5 Pa ( we convert standard pressure to Pa unit)

T1 = 427 K

P2 = 1.20 x 10^7

T2 = 429 K

delta H fus is enthalpy of fusion which we have to find out

delta V is volume change at melting point

Let us find delta V first

Let's say we have 1 mol of substance which would be 0.122 kg

volume of this substance in solid state would be

density (solid) = mass / volume (solid)

1.08 x 10^3 Kg/ m^3 = 0.122kg / volume (solid))

volume (solid) = 0.122 kg/ 1.08x 10^3 kg/m3

volume (solid) = 1.13 x 10^-4 m^3

similarly volume (liquid) = 0.122 kg/ 1.01 x 10^3 kg/m^3

volume (liquid) = 1.21 x 10^-4 m^3

delta V = 1.21 x 10^-4 m^3 - 1.13 x 10^-4 m^3

delta V = 8 x 10^-6 m^3

Let us substitute the values in Clausius Clapeyron equation

P2 - P1 = delta H (fus) / delta V * ( ln T2 - ln T1)

1.20 x 10^7 Pa - 1.01 x 10^5 Pa = delta H (fus) / 8 x 10^-6 m^3 * ( ln 429 - ln 427)

1.19 x10^7 Pa = delta H (fus) / 8 x 10^-6 m^3 * ( 4.67 x 10^-3 )

Delta H (fus) = 1.19 x 10^7Pa

------------------------- * 8 x 10^-6 m^3 = 20385 Pa*m^3

4.67 x 10^-3

Delta H (fus) = 20385 Pa*m^3

Pa* m^3 can be expressed in terms of Joules.

1 Pa*m^3 = 1 J

therefore we have Delta H (fus) = 20385 J

We had taken 1 mol of substance , so delta H (fusion) = 20385 J/ 1 mol

Delta H (fusion) = 20385 J / mol

We want to find delta S(fusion) here .

Delta H (fusion) & delta S (fusion) can be related to each other by the equation

Delta H (fusion) = T delta S (fusion)

Given : Delta H (fus) = 20385 J/mol

T = 429 K

substituting in above equation, we get

20385 J/mol K = 429 K * delta S (fus)

delta S (fus) = 20385 J/mol / 429 K

delta S (fus) = 47.5 J/mol K

b. We will again use Clausius Clapeyron equation which is given as

ln P2 - ln P1 = delta H (vap) /R * ( 1/T1 - 1/T2)

Given : P2 = 2.00 bar

P1 = standard pressure which is 1.01 bar

T1 = 550 ( standard boiling point)

T2 = ?

Delta H vap = 2.20 x 10^4 J/mol

Substituting the values, we get

ln 2 - ln 1.01 = 2.20 x 10^4 J/mol / 8.314 J/mol K * ( 1/550 - 1/T2)

0.683 = 2646 ( 1.818 x 10^-3 - 1/T2)

2.58 x 10^-4 = 1.818 x 10^-3 - 1/T2

1/T2 = 1.818 x 10^-3 - 2.58 x 10^-4 = 1.56 x 10^-3

T2 = 1/1.56 x 10^-3

T2 = 641 K

c. To solve for this, Kb of the solvent needs to be known. Please provide the data for the same.

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