Advertisements

Welcome to My Chemistry Tutor

Welcome, Guest. Please login or register.
Did you miss your activation email?

Login with username, password and session length

Search for Answers

Custom Search
Use our free chemistry search to search thousands of chemistry homework answers and tutorials

Ask a Question

If you did not find the answer to your chemistry question, join the community and ask for help.

Help Other Students

If you received chemistry help here why not try to help other students. You might undertand something that your peers are having trouble with. If you are in college chemistry you can likely help a high school chemistry student

How to calculate molality of a saline solution that contains 175.5 g of NaCl / L

Pages: [1]
  Print  
Author Topic: How to calculate molality of a saline solution that contains 175.5 g of NaCl / L  (Read 474 times)
farisallil
Labrat
*
Posts: 3




« on: October 25, 2009, 11:27:56 AM »

A saline solution contains 175.5 g of NaCl per liter of solution. Using a water density of 0.9982 g/cm^3, express the concentration of NaCl in term of molality.

I know that molality = moles of solute/ # of kilograms of solvent. But how do I get the number of kilograms of solvent from this problem?


Logged
kingchemist
Global Moderator
Hero Member
*****
Posts: 2880

A Liverpool FC fan




« Reply #1 on: October 25, 2009, 12:54:35 PM »
1

Mass of the solution (water + salt) = Density x Volume = 0.9982 x 1000 = 998.2 g

Mass of water = (998.2 - 175.5)g

I think you should be OK from here
hey
Logged

'Chemistry is not just the study of matter; Chemistry is the study which matters!' - Kingchemist
farisallil
Labrat
*
Posts: 3




« Reply #2 on: October 25, 2009, 02:16:21 PM »
2

Mass of the solution (water + salt) = Density x Volume = 0.9982 x 1000 = 998.2 g

Mass of water = (998.2 - 175.5)g

I think you should be OK from here

Well, 0.9882 is the density of pure water not the solution
hey
Logged
kingchemist
Global Moderator
Hero Member
*****
Posts: 2880

A Liverpool FC fan




« Reply #3 on: October 25, 2009, 02:50:07 PM »
3

OK, I misread the question. However, you will need to be given the density of the solution to work this out.
hey
Logged

'Chemistry is not just the study of matter; Chemistry is the study which matters!' - Kingchemist
farisallil
Labrat
*
Posts: 3




« Reply #4 on: October 25, 2009, 10:03:21 PM »
4

ok
I have looked into the textbook and found the attached table for densities of NaCl solutions
Here is my try after I found the table:
175.5 g/L = 1.4646 lb/gal
(1.587-1.356)/(1.1162-1.1009)=(1.587-1.464)/(1.1162-x)
--> specific gravity = x= 1.10809
specific gravity * density of water = density of solution
--> 1.10809*0.9982 = 1.1061 g/cm^3

density of solution * volume = 1.1061*1000 = 1106.1
mass of water = 1106.1-175.5=930.6 grams=0.9306 kg

molality = (175.5/58.43)/(0.9306)= 3.228 molal

However, the textbook says that the final answer is 3.198
so what did I do wrong?
hey


* table.jpg (88.8 KB, 713x343 - viewed 3 times.)
Logged
Pages: [1]
  Print  

 
Jump to:  

* Share this topic...
In a forum
(BBCode)
In a site/blog
(HTML)

Powered by MySQL Powered by PHP Powered by SMF 1.1.11 | SMF © 2006-2009, Simple Machines LLC Valid XHTML 1.0! Valid CSS!
MyChemistryTutor.com

Idea by BJR Technologies