Question

The boiling point of pure water is ${\text{373}}$ K. Calculate the boiling point of an aqueous solution containing ${\text{18}}$ g of glucose (MW=${\text{180}}$) in ${\text{100}}$ g of water. Molar elevation constant of water is $0 \cdot 52$K kg/mol.

Answer 1

Hint: Elevation in boiling point is a colligative property. Colligative properties are directly proportional to the number of particles of solute. When a solute is added to water, boiling point of water increases. This is called elevation in boiling point.

Complete step by step answer:
Elevation in boiling point is directly proportional to the number of particles of the solute in solution.
Hence we can write, $\vartriangle {T_b} = i{k_b}m$
Where,
$\vartriangle {T_b}$= elevation in boiling point
$i$ = vant-Hoff factor
${k_b}$= molal elevation constant
m = molality of solution
Pure water freezes at ${\text{373}}$ K. But since our solution contains glucose, the boiling point will be elevated from the normal value. We need to find this new boiling point. For that we need to calculate $\vartriangle {T_b}$.
Given ${k_b}$ of water=${\text{0}}{\text{.52}}$K kg/mol
i for glucose is equal to one.
Molality can be calculated by,
Molality (m) = $\dfrac{{{w_B} \times 1000}}{{{M_B} \times {W_A}}}$
Where,
${w_B}$ = weight of solute in grams
${M_B}$ = molecular weight of solute
${W_A}$ = weight of solvent in grams
Given,
${w_B} = 18g$
${M_B} = 180$
${W_A} = 100g$
Let us substitute these values into the equation of molality.
Molality (m) = $\dfrac{{18 \times 1000}}{{180 \times 100}} = 1$molal
Now substitute the value of molality into the equation, $\vartriangle {T_b} = i{k_b}m$.
$\vartriangle {T_b} = 1 \times 0 \cdot 52 \times 1 = 0.52$ K
Elevation in boiling point is $0.52$ Kelvin. Then the boiling point of solution will be,
T = Boiling point of pure water + elevation in boiling point
$T = 373 + 0.52 = 373.52$ K
Hence, the boiling point of an aqueous solution containing ${\text{18}}$ g of glucose in ${\text{100}}$ g of water is $373.52$ Kelvin.

Note:
The value of the van't Hoff factor, i depends on the nature of the solute and solvent. Glucose does not undergo association or dissociation when dissolved in water. Hence i for glucose is equal to one. But some solutes when dissolved in certain solvents undergo association /dissociation. In that case we need additional factors like degree of association/dissociation and number of particles into which one molecule associate/dissociate to calculate i.