Posted in the Pocahontas Forum
#1 Apr 28, 2009
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More Chemistry Homework help! >.<?
Can you explain how you get the answers so I get how to do other problems:
1) A solution consists of 10.3 g of the nonelectrolyte glucose, C6H12O6, dissolved in 250. g of water. What is the freezing-point depression of the solution?
2) In a laboratory experiment, the freezing point of an aqueous solution of glucose is found to be -0.325 degrees Celsius. What is the molal concentration of this solution?
3) If 0.500 mol of a nonelectrolyte solute are dissolved in 500.0 g of ether, what is the freezing point of the solution?
4) A solution contains 50.0 g of sucrose, C12H22O11, a nonelectrolyte, dissolved in 500.0 g of water. What is the boiling-point elevation?
5) A solution contains 450.0g of sucrose, C12H22O11,a nonelectrolyte, dissolved in 250 g of water. What is the boiling point of the solution?
6) If the boiling point elevation of an aqueous solution containing a nonvolatile electrolyte is 1.02 degrees Celsius, what is the molality of the solution?
Thanks for your help!
#2 Apr 28, 2009
No problem for the help.
Can you believe I actually had to have a minor in Chemistry to get a B.S.?
1) Zero degrees Celsius, which depresses me.
2).325 molal concetratoin X c squared, unless it happens on Wednesday.
4) The "boiling point evaluation" of the proposed problem is "that is a very good boiling point".
5)100 degree Celsius.
6) Three megahertz!
#3 Apr 28, 2009
get 12 ounces dihydrogenoxide,
6 ounces whiskey.
mix into a quart jar full of ice,
consume before ice gets to 33 degress F.
#4 Apr 28, 2009
#5 Jun 29, 2009
Hi Lindsey, Here I will help you with the first one, and see how you do.
First you need to find the molality which is mol / kg
10.3 g * 1 mol / 180.156 g = 0.0572 mol Glu
molality = 0.0572 mol Glu / 0.250 kg = 0.2287 m
Now for the freezing point depression
∆Tf = Kf * m
And the boiling point elevation
∆Tb = Kb * m
where Kb for water is 0.521 °C / m
Where Kf for water is 1.858 °C / m
∆Tb = 0.521 °C /m * 0.2287 m = 0.119
∆Tf = 1.858 °C / m * 0.2287 m = 0.425
So the boiling point is now 100 °C + 0.119 °C = 100.119 °C
freezing point is 0°C - 0.425 °C =-0.425 °C
Now for the rest of the problems
2) do this backwards
3) do the same I did for #1 but use Kb = 2.20 Kf = 7.88
4) follow what I did
5) follow what I did
6) solve for m
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