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
# 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.
Thank you
so then would the new points be:
b.p. = 218.67 C
f.p. = 16.38 C ?
To solve this problem, you need to use colligative properties, specifically the freezing point depression and boiling point elevation formulas. From your calculations, it seems like you have followed the correct steps.
To find the change in boiling point, you used the formula:
boiling point = 100 degrees C + deltaTbp
deltaTbp represents the change in boiling point, which can be calculated using the equation:
deltaTbp = molality × Kbp
In this case, the molality is 0.056 mol solution, and Kbp for acetic acid is 3.07 C/m. So, you multiplied the molality by Kbp to get the change in boiling point, and added it to the normal boiling point of 100 degrees C.
Similarly, to find the change in freezing point, you used the formula:
freezing point = 0 degrees C - deltaTfp
deltaTfp represents the change in freezing point, which can be calculated using the equation:
deltaTfp = molality × Kfp
In this case, the molality is 0.056 mol solution, and Kfp for acetic acid is 3.90 C/m. So, you multiplied the molality by Kfp to get the change in freezing point, and subtracted it from the normal freezing point of 0 degrees C.
According to your calculations, the new boiling point is approximately 100.173 degrees C, and the new freezing point is approximately -0.218 degrees C.
To confirm if your answers are correct, you can compare them with the expected values or consult your textbook or teacher for confirmation. If you are still unsure, you may want to double-check your calculations and make sure you are using the correct values for Kf and Kb for acetic acid.