× Didn't find what you were looking for? Ask a question
Top Posters
Since Sunday
r
4
L
4
3
d
3
M
3
l
3
V
3
s
3
d
3
a
3
g
3
j
3
Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
New Topic  
RiverTmasco RiverTmasco
wrote...
Posts: 95
Rep: 0 0
11 years ago
Would a 1m aqueous solution of glucose, C6H1206, or a 1m aqueous solution of the ionic compound,(NH4)3 PO4, have a lower freezing point? Explain your reasoning and prove it with Tf calculations. Kf for water is 1.86 degrees Celcius/m.

Thank you.
Read 634 times
3 Replies

Related Topics

Replies
wrote...
#1
11 years ago
delT = Kf*m*i.

i = 1 for glucose, and i > 1 for ammonium phosphate (depending on how much it dissociates, if it's completely, n = 4 I think). Plugging in a higher value for i will result in a larger delta T, so the freezing point will change more.
wrote...
#2
11 years ago
NH4)3 PO4 since that would dissociate into 4 m.

Tf = i*m*Kf

i = van hoffs constant or how much something dissociates

For NH4)3 PO4
i = 4

since it dissociates into 3 NH4 + n PO4 3-

So 4*1*1.86 = Tf

then subtract 0 - Tb to get freezing point for NH4)3 PO4

For molecular compound i = 1

so Tf = 1*1*1.86 = 1.86

0-1.86 = -1.86

The freezing pt for NH4)3 PO4 should be 4*-1.86
wrote...
#3
11 years ago
lowering of freezing point is a colligative property. Colligative property depends upon no. of particles . so more the no. of particles of solute more the property. Clearly glucose ( as it don't dissociates) will have lesser no. of paticles than 1m ionic compound, so ionic compound will have lower freezing point than 1m glucose.
New Topic      
Quick Reply
Explore
Post your homework questions and get free online help from our incredible volunteers
  1426 People Browsing
Start New Topic Take the Tour CoursesNew Study Tools Topics Trending Browse by Textbook
Live Chat
Related Images
  
 308
  
 263
  
 251