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

- Chemistry -
**DrBob222**, Friday, October 22, 2010 at 12:19pm
# of moles

(5g C6H12O6)(1mol C6H12O6/ 180g C6H12O6)

= 0.028 mol of C6H12O6

** Your calculations are ok but you didn't finish. After your have delta T you must add to normal boiling point and subtract from normal freezing point. Also, I think you transferred f.p. and b.p. incorrectly from the problem. I don't like to round at each step as I go through a problem; rounding at each step can cause rounding errors at the end. I prefer to just leave those "extra" numbers in the calculator. For example:
**

(5.0/180)/0.5 x 3.90 = 0.21667 which rounds to two places as 0.22 (if 5.0 is the correct value from your problem then two s.f. is all you can have). Then -0.22+16.6 = ??

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

**I think the problem quotes normal boiling point as 118.5**

freezing point = 0 degreesC-deltaTfp

= 0 degreesC-mKfp

= O degreesC-(0.056m)(3.90C/m)

= -0.218 degrees C

**I think the problem quotes normal freezing point as 16.6.**

- Chemistry -
**Angelina**, Friday, October 22, 2010 at 1:03pm
Thank you

so then would the new points be:

b.p. = 218.67 C

f.p. = 16.38 C ?

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