Monday
August 31, 2015

Homework Help: Chemistry

Posted by Angelina on Friday, October 22, 2010 at 2:48am.

I'm not sure if i did this problem right or not. I keep reading through my book, and i think i followed the equations right, but i'm not confident with my answers.

The problem:

5.0 g of glucose,C6H12O6, is disolved in 500.0 g of acetic acid. What is the new freezing point and boiling point for the solution? Kf acetic acid = 3.90, Kb acetic acid = 3.07 (normal freezing point for acetic acid = 16.60 degrees C, boiling point = 118.5 degrees C)

my work:

# of moles
(5g C6H12O6)(1mol C6H12O6/ 180g C6H12O6)
= 0.028 mol of C6H12O6

molality
(0.028mol/500g CH3COOH)(1000g CH3COOH/1kg)
= 0.056 mol solution

boiling point = 100degreesC+deltaTbp
= 100degreesC+mKbp
= 100degreesC+(0.056m)(3.07 C/m)
= 100.173 degrees C

freezing point = 0 degreesC-deltaTfp
= 0 degreesC-mKfp
= O degreesC-(0.056m)(3.90C/m)
= -0.218 degrees C

now, is that all there is (IF this is correct...)or is there more to solving this equation.

I really need help..Im so confused...Thank you

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