In making candy, a certain recipe calls for heating an aqueous sucrose solution to the "softball" stage. Which has a boiling point of 235-240 F. Wat is the range of mass percentages of the solution of sugar (C12H22O12) that boils at those two temperature?

I'll start by converting degrees Fahrenheit to degrees Celsius.

`"235 F" = "112.78"^@"C"`
`"240 F" = "115.56"^@"C"`

The equation for boiling point elevation is

`DeltaT_b = i * K_b * b`, where

`DeltaT_b` - the poiling point elevation;
`i` - the van't Hoff factor - in your case `"i=1"` because sucrose does not dissociate when dissolved in water;
`K_b` - the ebullioscopic constant - for water its value is listed as `"0.512 "^@"C" * "kg/mol"`;
`b` - the molality of the solution.

Water's normal boiling point is `"100"^@"C"`, which means that the boiling point elevation for each solution will be

`DeltaT_(b1) = "112.78"^@"C" - "100.0"^@"C" = "12.78"^@"C"`
`DeltaT_(b2) = "115.56"^@"C" - "100.0"^@"C" = "15.56"^@"C"`

Let's start by determining the molalities of the two solutions

`b_1 = (DeltaT_(b1))/(K_b) = ("12.78"^@"C")/("0.512 "^@"C" * "kg/mol") = "25.0 mol/kg"`

`b_2 = (DeltaT_(b2))/(K_b) = ("15.56"^@"C")/("0.512 "^@"C" * "kg/mol") = "30.4 mol/kg"`

The first solution will have `"25.0 moles"` of sucrose for every `"1 kg"` of water. For simplicity, let's assume we have `"1 kg"` of water. This means that the mass of sucrose in solution will be

`"25.0 moles" * ("342.3 g")/("1 mole") = "8557.5 g"`

Percent concentration by mass is defined as the mass of the solute divided by the total mass of the solution and multiplied by 100. Since we have `"1 kg = 1000 g"` of water, the first solution's concentration by mass is

`"%m" = ("8557.5 g")/("8557.5 g + 1000 g") * 100 = 89.5%`

Likewsie, the second solution will have `"30.4 moles"` of sucrose for every `"1 kg"` of water.

`"30.4 moles" * ("342.3 g")/("1 mole") = "10406 g"`

This solution's percent concentration by mass will be

`"%m" = ("10406 g")/("11406 g") * 100 = 91.2%`