[mikepl1]

Could anyone help me with these two problems? The answer for the first one is 89.5%- 91.2% and the second one is 213 grams of glucose, but I didn't get those answers.

1. In making candy, a certain recipe calls for heating an aqueous sucrose solution to the "soft-ball" stage which has a boiling point of 235-240 degrees Fahrenheit. What is the range of mass percentages of the solutions of sugar (C12H22O11) that boil at those two temperatures? (Hint: Remember that temperature changes are in Celsius)
2. How many grams of glucose (molar mass= 180.9 g/mol) must be dissolved in 255 g of water to raise the boiling point to 102.36 deg. Celsius?

[Sophia7X]

Are you using the difference in temperature, rather than just using the number (ie 102.36 as opposed to 2.36 degrees Celsius since a boiling point of 102.36 means that the boiling point of the solution has been elevated 2.36 degrees compared to pure water)?

[Stovn0611]

Use the equation ΔT_b=K_b*m*i

ΔT_b is the change in boiling point temperature
K_b is a constant and its value for water is 0.51
m= molality
i= the number of moles of solute formed per mole of substance dissolved

1. 235-240 degrees Fahrenheit= 112.778 degrees Celsius - 115.556 degrees Celsius

The normal boiling point of water is 100 degrees Celsius so

ΔT_b=K_b*m*i
(112.778-100)=0.51*m*1  (sucrose only creates 1 mole of solute when dissolved)
m = 25.055 mol/kg solvent
To get mass percentage:
Assume there's 1 kg solvent
There will be 25.055 mol of sucrose which is 25.055 mol*.3422965 kg/mol = 8.5762388075 kg

Mass percentage of sucrose in solution = 8.5762388075/(8.5762388075+1)≈89.5%

I'll let you do the calculations for 240 degrees Fahrenheit

2. Once again, use the equation:

ΔT_b=K_b*m*i

Edit by Borek: Solve for m.

Then, just use various conversion factors to solve for the grams of glucose in the water