# 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?

Jan 30, 2015

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

$\text{235 F" = "112.78"^@"C}$
$\text{240 F" = "115.56"^@"C}$

The equation for boiling point elevation is

$\Delta {T}_{b} = i \cdot {K}_{b} \cdot b$, where

$\Delta {T}_{b}$ - the poiling point elevation;
$i$ - the van't Hoff factor - in your case $\text{i=1}$ because sucrose does not dissociate when dissolved in water;
${K}_{b}$ - the ebullioscopic constant - for water its value is listed as $\text{0.512 "^@"C" * "kg/mol}$;
$b$ - the molality of the solution.

Water's normal boiling point is $\text{100"^@"C}$, which means that the boiling point elevation for each solution will be

$\Delta {T}_{b 1} = \text{112.78"^@"C" - "100.0"^@"C" = "12.78"^@"C}$
$\Delta {T}_{b 2} = \text{115.56"^@"C" - "100.0"^@"C" = "15.56"^@"C}$

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 $\text{25.0 moles}$ of sucrose for every $\text{1 kg}$ of water. For simplicity, let's assume we have $\text{1 kg}$ of water. This means that the mass of sucrose in solution will be

$\text{25.0 moles" * ("342.3 g")/("1 mole") = "8557.5 g}$

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 $\text{1 kg = 1000 g}$ 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 $\text{30.4 moles}$ of sucrose for every $\text{1 kg}$ of water.

$\text{30.4 moles" * ("342.3 g")/("1 mole") = "10406 g}$

This solution's percent concentration by mass will be

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