Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Out of questions for the month can you answer both please Thank you The normal b

ID: 995028 • Letter: O

Question

Out of questions for the month can you answer both please Thank you

The normal boiling point of liquid propanol is 371 K. Assuming that its molar heat of vaporization is constant at 43.9 kJ/mol, the boiling point of C3H7OH when the external pressure is 0.625 atm is K. step by step and formula please thank you

Consider the following aqueous solution

0.15 m CaCl2

0.25 m CsI

0.20 m caffeine (a nonelectrolyte)

0.10 m K3PO4

Which solution has the highest freezing point?

Which solution has the lowest vapor pressure of water?

Explanation / Answer

Colligative property is dependent on no of solute particles i.e. no of ions/species.

Depression in freezing point and Relative lowering of vapor pressure are colligative properties.

Species with lowest value of (molality x no of ions/species) has lowest Depression in freezing point which in turn has highest freezing point.

Species with highest value of (molality x no of ions/species) has highest Relative lowering of vapor pressure  which in turn has lowest vapor pressure of water.

a) 0.15 m CaCl2

= 0.15 x 3 ( CaCl2 = Ca+2 + 2Cl- = 3 ions)

= 0.45

b) 0.25 m CsI

= 0.25 x 2 ( CsI = Cs+ + I- = 2 ions)

= 0.50

c) 0.20 m caffeine (a nonelectrolyte)

= 0.20 x 1

= 0.20

d) 0.10 m K3PO4

= 0.10 x 4 ( K3PO4 ------------> 3K+ + PO43-)

= 0.4

Therefore,

0.20 m caffeine has highest freezing point.

0.25 m CsI has the lowest vapor pressure of water.

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote