3 qs on destroyer

[gomawum]

---

Members don't see this ad.

Q.why is Pyrrole poor base?

even though the it lose hydrogen from the nitrogen, it still has one pair of electrons in the same plane whereas another not in plane, isn't it?




Q.which solution has the lowest freezing point?

1.4M Na2So4 (i=3)
2.4M C2H5OH (i=2)

even though it's colligative law, freezing depression depends on molality, constant, but how do u know 1 has lower freezing point than 2?



Q.C4H10 has more entropy than C3H8

why is that? I see C4H10 has more complexity so occupy more space and will move slowly which decrease entropy.

 

[rose786]

For your 2nd question, read the answer explanation. It tells you how to calculate something called the "effective molar concentration.

For the 3rd question, take a look at what phases both of those compounds are in (solid vs gas vs liquid). I'm guessing that is the reason for the greater entropy.

 

[gomawum]

for 2nd one if we convert molarity to mole and assume that solvent has same mass
than molality is proportional to mole so it convert to

1.4M Na2So4 (i=3) -> 1.4 x 3 = 4.2
2.4M C2H5OH (i=2) -> 2.4 x 2 = 4.8

so isn't 2nd one has lower freezing point?


3rd question doesn't state about the phases, it just says that c4h10 has more atoms, but since we are looking at the molecule, not atoms, I don't know how it relates to the entropy

 

[Orgodox]

Q.why is Pyrrole poor base?

Even though the it lose hydrogen from the nitrogen, it still has one pair of electrons in the same plane whereas another not in plane, isn't it?

If the lone pair on the Nitrogen is used as a base and forms another N-H bond, it will lose aromaticity. Pyrrole has 6 e's and that satisfies Huckles rule but if it uses the lone pair it will only have 4 e's and wont be fully conjugate which violates 2 of the rules for aromaticity.