Place the following substances in order of increasing vapor pressure at a given temperature. a. NF3,
b. NH3,
c. BCl3.

The trend for increasing vapor pressure will be:

NH₃<NF₃<BCl₃

We know that,

The stronger the intermolecular forces, the lower the vapor pressure, and the higher the boiling point.

Thus to solve this question we need to look the intermolecular attractions in the given molecules.

1. In ammonia molecule, since it is polar in nature. It exhibits, dipole-dipole interactions and also there is hydrogen bonding between nitrogen and hydrogen atoms. So these interactions are strongest  thus NH₃ will have lower value of vapor pressure.

2. Forces of attractions in NF₃ is also stronger in comparison to BCl₃ but weaker than NH₃ thus in terms of vapor pressure it will come after NH₃.

3. Lastly, in BCl₃ it is a non-polar molecule; it exhibits London forces which are the weakest thus it will have the higher value of vapor pressure.

Trend for intermolecular attraction:

ion-dipole>hydrogen bonding>dipole-dipole> London forces > Van der Waals forces

Therefore,

Increasing order of vapor pressure will be:

NH₃<NF₃<BCl₃

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NH3 < NF3 < BCl3

Explanation:

The vapour pressure of a substance has something to do with the nature of intermolecular forces between its molecules. If the molecules of a substance are held together by strong intermolecular forces, the substance will display a low vapour pressure at a given temperature and vice versa.

Ammonia has the lowest vapour pressure because of strong intermolecular hydrogen bonds that hold its molecules together.

58750.08068 is the answer i think