Answer:
Boiling point of solution is 
Explanation:
Cane sugar is a non-volatile solute.
According to Raoult's law for a non-volatile solute dissolved in a solution-
Where, 
is elivation in boiling point of solution, 
is ebbulioscopic constant of solvent (how much temperature is raised for dissolution of 1 mol of non-volatile solute) and m is molality of solution.
Here, 
610 g of cane sugar = 
moles of cane sugar
= 1.78 moles of cane sugar
So, molality of solution (m) = 
Plug in all the values in the above equation, we get-
So, boiling point of solution = 
1) All the CO2 comes from the C24 H30 N3 O
2) To balance C from CO2 with C from C24, the ratio of cC24 H30 N3 O to CO2 is 24 / 1
3) Convert 1.20 g of CO2 to number of moles
number of moles = mass in grams / molar mass
mass = 1.20 g
molar mass = 12.01 g/mol + 32.00 g/mol = 44.01 g/mol
number of moles = 1.2 g / 44.01 g/mol = 0.0273 mol
4) proportion
1 mol C24 H30 N3 O / 24 mol CO2 = x / 0.0273 mol CO2
=> x = 0.0273 mol CO2 * 1 mol C24H30N3O / 24 mol CO2 = 0.00114 mol C24H30N3O
5) Convert 0.00114 mol C24H30N3O to grams
molar mass = 376.5 g/mol
mass = 0.00114 mol * 376.5 g/mol = 0.429 g
6) mass percent in the mixture
mass percent = (mass of lsd / mass of mixture) * 100 = (0.429g / 1.0g) * 100 = 42.9%
Answer: 42.9%
