How many grams of glucose (molar mass = 180.9g/mol) must be dissolved in 255g of water to raise the boiling point to 102.36Celsius?

Jan 30, 2015

The answer is $\text{213.5 g}$ of glucose.

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

$\Delta {T}_{b}$ - the poiling point elevation;
$i$ - the van't Hoff factor - in your case $\text{i=1}$ because glucose 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.

You know that pure water boils at $\text{100.0"^@"C}$, which means that the boiling point elevation for this solution is

$\text{102.36"^@"C" - "100.0"^@"C" = "2.36"^@"C}$

This means that the solution's molality is

b = (DeltaT_b)/K_b = ("2.36 "^@"C")/("0.512 "^@"C" * "kg/mol") = "4.61 mol/kg"

This means that you have $\text{4.61 moles}$ of glucose for every $\text{1 kg}$ of water. The number of moles you'll have in $\text{255 g}$ of water will be

$\text{255 g" * ("4.61 moles")/("1000 g") = "1.18 moles}$

Since you know glucose's molar mass, the mass of glucose will be equal to

$\text{1.18 moles" * ("180.9 g")/("1 mole") = "213.5 g}$, or

m_("glucose") = "214 g" - rounded to three sig figs.