# 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?

##### 1 Answer
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%