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True [87]
3 years ago
12

Positive Deviation from Raoult's Law occurs when the vapour pressure of component is greater than what is expected in Raoult's L

aw. For Example, consider two components A and B to form non-ideal solutions. Let the vapour pressure, pure vapour pressure and mole fraction of component A be PA, PAo and XA respectively and that of component B be Ps, PB° and xB respectively. These liquids will show positive deviation when Raoult's Law when: a. PIA]> P[AO] times x[A] and PIB] > PIBO] times xIB], as the total vapour pressure (PIAO] XIA] + P[B0] x[B]) is greater than what it should be according to Raoult's Law. O b. The solute-solvent forces of attraction is weaker than solute-solute and solvent-solvent interaction O c. The enthalpy of mixing is positive because weaker binding forces or even repulsion are resulted O d. The volume of mixing is positive as weaker binding forces have led to an expansion in volume O a and b only O a, b and c only O All of the above
Chemistry
1 answer:
8090 [49]3 years ago
6 0

Answer:

All of the above.

Explanation:

In positive deviation from Raoult's  Law occur when the vapour pressure of components is greater than what is expected value in Raoult's law.

When a solution is non ideal then it shows positive or negative deviation.

Let two solutions A and B to form non- ideal solutions.let the vapour pressure  of component A is P_A and vapour pressure of component B is P_B.

P^0_A= Vapour pressure of component A in pure form

P^0_B= Vapour pressure of component B in pure form

x_A=Mole fraction of component A

x_B==Mole fraction of component B

The interaction between A- B is less than the interaction A- A and B-B interaction.Therefore, the escaping tendency of liquid molecules in mixture is greater than the escaping tendency in pure form.Hence, the vapour pressure of a mixture is greater than the initial value of vapour pressure.

P_A >P^0_A\cdot x_A,P_B>P^0_B\cdot x_B

Therefore, P_A+P_B >P^0_A\cdot x_A+P^0_B \cdot x_B

Therefore, the enthalpy of mixing is greater than zero and change in volume is greater than zero.

Hence, option a,b,c and d are true.

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The question is incomplete. The complete question is :

The table compares the number of electrons in two unknown neutral atoms.

Comparison of Electrons

Atom Number of Electrons

A 9

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Use this information to determine the number of valence electrons in the atoms. Which of the following correctly compares the stability of the two atoms?

A. Both are unreactive.

B. Both are highly reactive.

C. A is unreactive and D is reactive.

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Answer:

B. Both are highly reactive.

Explanation:

Atomic number of electrons given is 9 and 11, it means that atom A is fluorine (F) and atom D is Sodium (Na).

In order to gain stability 8 electrons are required in outer shell. So atoms either gain or lose electrons in its outer shell to become more reactive.

Sodium has electronic configuration (2,8,1)  and easily loses one electron and become  (2, 8) which is highly reactive.

Fluorine has electronic configuration (2, 7) and can gain easily an electron and becomes reactive to form (2, 8).

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