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Last updated Jun 29, 2009
#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 freezingpoint 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 boilingpoint 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.? Answers, 1) Zero degrees Celsius, which depresses me. 2).325 molal concetratoin X c squared, unless it happens on Wednesday. 3)Absolute Kelvin. 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
Too Funny!! 

#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 Simple. 

 
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