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**Hint:**the fact that at boiling point of the liquid, liquid & its vapor are in equilibrium i.e. vapor is in equilibrium at one atmospheric pressure. We know that at equilibrium, Gibbs energy change is zero. ie $\Delta G = 0$.

**Complete step by step solution:**

It is given that Enthalpy of vaporization is 30 Kilojoules per mole.

i.e. $\Delta H$ = enthalpy of vaporization = 30000 $Jmo{l^{ - 1}}$

It is also given that Entropy of vaporization is 75$Jmo{l^{ - 1}}{K^{ - 1}}$

i.e. $\Delta S$ = entropy of vaporization = 75 $Jmo{l^{ - 1}}{K^{ - 1}}$

We know at boiling point of the liquid, vapor is in equilibrium (at one atmospheric pressure) which implies that,

Gibbs energy change is equal to zero.

i.e $\Delta G = 0$

We know that,

$\Delta G = \Delta H - T\Delta S$

Substituting the values in the equation we get,

0 = 30000 – 75T

$\therefore T = \dfrac{{3000Jmo{l^{ - 1}}}}{{75Jmo{l^{ - 1}}{K^{ - 1}}}}$

Therefore, we get;

T = 400 K

∴ Boiling point of the liquid at one atmospheric pressure is 400k.

∴ Correct option - B. 400k.

**Additional information:**

The change in free energy occurs when a compound is formed from its elements in their most thermodynamically stable state at standard state conditions i.e. 1 atm.In thermodynamics, the Gibbs free energy is a thermodynamic potential that can be used to calculate the maximum of irreversible work that may be performed by a thermodynamic system at a constant temperature and pressure.The Gibbs free energy is the maximum amount of non-expansion work that can be extracted from a thermodynamic closed system. This maximum can be attained only in a completely reversible process.

**Note:**Enthalpy of vaporization is the amount of energy that must be added to the liquid substance, to transform a quantity of that substance into gas. Entropy of vaporization is an increase in entropy upon vaporization of a liquid.

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