home > q&a board > chemistry > topic: molality

Q BgQuestion:

Scholar
Karma Points: 200
Respect (92%):
posted by  jude07 on 9/5/2008 10:17:07 AM  |  status: Closed  

Molality

Course Textbook Chapter Problem
N/A N/A N/A N/A
Question Details:
How do you calculate the molality of the following aqueous solution:  48.2 percent by mass KBr solution.

AAnswers:

Answer Question
Scholar
Karma Points: 232
posted by sabrewulf on 9/5/2008 10:27:59 AM  |  status: Live
Asker's Rating: Helpful   
jude07's comment:
"Thank YOU"
Response Details:
Molality is moles of solute per kilogram of solvent.  So, assume that the solution is 100 kg.  Then, 51.8 kg would be the solvent.  Since the solvent is not specifically stated, it is assumed to be water.

Now, for the moles of solute.  48.2 kg KBr = (48.2)(1 mol/0.119 kg) = 405.042 mol KBr

Finally, use the molality formula.   405.042 mol KBr/ 51.8 kg solvent = 7.82 molal

Hope this helped.
Scholar
Karma Points: 200
posted by jude07 on 9/5/2008 10:52:16 AM  |  status: Live
Asker's Rating: N/A-Posted by Person Asking Question   
Response Details:
Where doe you get the 51.8kg?
(Cramster SME)
Moderator
posted by Werner on 9/5/2008 11:12:41 AM  |  status: Live
Asker's Rating: Lifesaver   
jude07's comment:
"Thank You very Much, this help a lot!"
Response Details:
  Given that
  The percent of KBr = 48.2 %
  That means 100 g of solution contains 48.2 g KBr and 51.8 g solvent
   The number of moles of KBr = mass of KBr / Molar mass of KBr
                                               = 48.2 g / 119 g/mol
                                               = 0.405 mol
    The mass of solvent = 51.8 g
                                   = 0.0518 kg
   Molality of solution = moles of solute / mass of solvent in kg
                                 = 0.405 mol / 0.0518 kg
                                 = 7.82 molal
 
   In first post 51.8 kg means if you take 100 kg of solution then the mass of solvent is 51.8 kg and the mass of solute is 42.2 kg.
 
Answer Question
Ask New Question

Join Cramster's Community

Cramster.com brings together students, educators and subject enthusiasts in an online study community. With around-the-clock expert help and a community of over 100,000 knowledgeable members, you can find the help you need, whenever you need it. Join for free today » How Cramster is different than tutoring »

Need Help In

Useful Links