Rank the given substances in order of decreasing vapor pressure.
SrCl2 (aq), NH3, CH4, Se2O
Place the following substances in order of decreasing vapor pressure at a
PF_5 BrF_3 CF_4
A) BrF_3 > PF_5 > CF_4
B) BrF_3 > CF_4 > PF_5
C) PF_5 > BrF_3 > CF_4
D) CF_4 > BrF_3 > PF_5
E) CF_4 > PF_5 > BrF_3
Place the following substances in order of increasing vapor pressure at NF_3 NH_3 BCI_3
A) NH_3 < NF_3 < BCI_3
B) NF_3 < NH_3 < BCI_3
C) BCI_3 < NF_3 < NH_3
D) NH_3 < BCI_3 < NF_3
E) BCl_3 < NH_3 < NF_3
Define critical point.
A) The temperature and pressure above which a supercritical fluid exist
B) The temperature and pressure where liquid, solid, and gas are equal equilibrium.
C) The temperature and pressure below which a supercritical fluid exist
D) The temperature and pressure are equal.
E) The temperature and pressure are not equal.
Answer to question 1
Higher intermolecular forces present within the compound will have low vapour pressure.
CH4 has very weak vanderwall force so it has very high vapour pressure
2-methylbutane has Vander wall force greater than the CH4 so it’s vapour pressure will be slightly less
Pentane has high Vander wall forces because structure is symmetrical so it has low vapour pressure.
Butanol has hydrogen bonding so it will have very low vapour pressure.
4 , 2 ,3, 1
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Answer to question 2
The given molecules are SrCl2(aq),NH3,CH4,Se2OSrCl2(aq),NH3,CH4,Se2O .
The intermolecular forces present between given molecules are as follows,
SrCl2(aq)SrCl2(aq)- ionic bond as there is large difference between both Sr and Cl atoms.
NH3NH3- covalent bond along with hydrogen bonding as there is electronegativity difference between N and H.
CH4CH4- covalent bond as both C and H shares electron to form compound.
Se2OSe2O- dipole-dipole forces.
The decreasing order of following given compounds is as follows,
Answer to question 3
The correct order of vapor pressure is
1>Option (E) follows the correct order of vapor pressure.
2>Option(C) Follows the correct order of vapor pressure.
3>Option(B) is the correct answer.
Higher is the molecular weight, more will be the van der wall force of attraction.
More is the van der wall force of attraction, higher will be the boiling point.
Higher is the boiling point, less amount of vapor is formed.
Less is the amount of vapor formed, lower will be the vapor pressure.