Raoult's Law, Finding an unknown compound

  • Thread starter Thread starter Glitche
  • Start date Start date
  • Tags Tags
    Law
Click For Summary
SUMMARY

The discussion centers on determining the identity of an unknown compound X in a 1.00m aqueous solution, which has a boiling point of 101.04°C. Using the boiling point elevation formula ΔTb = (Kb)(m), where Kb for water is 0.52, the change in boiling point is calculated as 1.04°C. This leads to the conclusion that two particles are present in the solution, suggesting that KCl (potassium chloride) is the most likely candidate for compound X due to its dissociation into two ions.

PREREQUISITES
  • Understanding of colligative properties, specifically boiling point elevation
  • Familiarity with the formula ΔTb = (Kb)(m)
  • Knowledge of ionic vs. molecular compounds
  • Basic chemistry concepts related to solutions and solubility
NEXT STEPS
  • Study the effects of ionic compounds on boiling point elevation
  • Learn about colligative properties in detail
  • Explore the dissociation of salts in aqueous solutions
  • Investigate other colligative properties such as freezing point depression
USEFUL FOR

Chemistry students, educators, and professionals involved in solution chemistry and colligative properties analysis.

Glitche
Messages
4
Reaction score
0
1. A 1.00m aqueos solution of compound X had a boiling point of 101.04 Celsius. Which one of the following could be compound X? The boiling point elevation constant for water is 0.52

A) CH3CH2OH
B) KCL
C) C6H12O6
D) NA3PO4




ΔTb=(Kb)(m)




1: identify the change in bp temperature. in this case, it's 101.04-100 = 1.04 degrees C.
2. divide this change by your bp elevation constant. 1.04/0.52 = 2

so now I think that means that two different molecules were added. My guess is it was KCl (potassium chloride). but of course, I'm not entirely sure =/

Can someone help me out?

 
Physics news on Phys.org
Looks OK to me.
 
Thanks Borek!
 

Similar threads

Replies
2
Views
17K
  • · Replies 18 ·
Replies
18
Views
10K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K