Top Posters
Since Sunday
5
a
5
k
5
c
5
B
5
l
5
C
4
s
4
a
4
t
4
i
4
r
4
New Topic  
wrote...
Educator
2 months ago
I am a bit confused. Isn't the value for molality mols solute/kg solvent? In this solution, you used kg of the solution not the solvent. How would you figure out the kg of ONLY the solvent?

Not sure I follow.

The question here is asking for "moles", not molality

"How many moles of AgCl are contained in 244 mL of 0.135 m AgCl solution?"
wrote...
2 months ago
I know that the question is asking for mols of AgCl, but it says that the solution is 0.135 m. Correct me if I'm wrong, but I was taught that the lowercase m represents molality.

Therefore, the solution is 0.135 moles AgCl/1 kg solvent (because molality is moles of solute/kg of solvent)

In the solution, they multiplied 0.135 m * 297.68 g, where 297.68 g is the kilograms of the SOLUTION.

The problem is that 0.135 m is equivalent to 0.135 moles AgCl/1 kg SOLVENT.

So, multiplying 0.135 by 0.29768 kg is multiplying  0.135 moles AgCl/1 kg SOLVENT *  0.29768 kg SOLUTION.

So, the units do not quite cancel out. I was thinking about how I can some how find the kilograms of solvent from the kilograms of solution, but I am not sure how to do it.
wrote...
Educator
2 months ago
I know what you mean, it is ambiguous, but in terms of this question the m stands for molarity, that is, mol/L. Otherwise you're right.

But technically it should be capital M
  New Topic      
Explore
Post your homework questions and get free online help from our incredible volunteers
  1364 People Browsing
Related Images
  
 96
  
 630
  
 182
Your Opinion
Which of the following is the best resource to supplement your studies:
Votes: 292