Question

A solution of water (kb=0.512 ∘c/m) and glucose boils at 101.56 ∘c. what is the molal concentration of glucose in this solution? assume that the boiling point of pure water is 100.00 ∘c.express your answer to three significant figures and include the appropriate units.

Answer 1

Answer is: molal concentration of glucose in this solution is 3,04 mol/kg (molal).
Kb(H₂O) = 0,512 °C/m.
T(glucose) = 101,56 °C.
T(H₂O) = 100°C.
ΔT = 101,56°C - 100°C.
ΔT = 1,56°C.
ΔT = m(glucose) · Kb(H₂O).
m(glucose) = ΔT ÷ Kb(H₂O).
m(glusose) = 1,56°C ÷ 0,512 °C/m.
m(glucose) = 3,04 mol/kg.
m - molal concentration.

Answer 2

The molal concentration of glucose in the given solution is $\boxed{{\text{3}}{\text{.05 m}}}$.

Further Explanation:

The properties that depend only on the concentration of solute are called colligative properties. These are listed below:

1. Relative lowering of vapor pressure.

2. Elevation in boiling point.

3. Depression in freezing point.

4. Osmotic pressure.

All these properties depend on the number of moles of solute particles and therefore called colligative properties.

The formula to calculate the elevation in boiling point is as follows:  $\Delta {{\text{T}}_{\text{b}}} = {{\text{k}}_{\text{b}}}{\text{m}}$   …… (1)                                                                                  

Where,

$\Delta {{\text{T}}_{\text{b}}}$ is the boiling point elevation.

${{\text{k}}_{\text{b}}}$ is the molal boiling point constant.

m is the molality of the solution.

Rearrange equation (1) to calculate the molality of the solution.

${\text{m}} = \dfrac{{\Delta {{\text{T}}_{\text{b}}}}}{{{{\text{k}}_{\text{b}}}}}$                                                                                                         …… (2)

The temperature change $\left( {\Delta {{\text{T}}_{\text{b}}}} \right)$ can be calculated as follows:

$\Delta {T_{\text{b}}} = {\text{Boiling}}\;{\text{point}}\;{\text{of }}{\text{glucose}} - {\text{Boiling}}\;{\text{point}}\;{\text{of }}{\text{pure water}}$        …… (3)                  

Substitute $101.56{\text{ }}^\circ {\text{C}}$ for the boiling point of glucose and $100{\text{ }}^\circ {\text{C}}$ for the boiling point of pure water in equation (3).

 $\begin{aligned}\Delta{{\text{T}}_{\text{b}}} &= 101.56{\text{ }}^\circ {\text{C}} - 100{\text{ }}^\circ {\text{C}}\\&= {\text{1}}{\text{.56 }}^\circ {\text{C}}\\\end{aligned}$

Substitute $1.56{\text{ }}^\circ {\text{C}}$ for $\Delta {{\text{T}}_{\text{b}}}$ and $0.512{\text{ }}^\circ {\text{C/m}}$ for ${{\text{k}}_{\text{b}}}$ in equation (2) to calculate the molality of glucose.

$\begin{aligned}{\text{m}} &= \left( {\frac{{1.56{\text{ }}^\circ {\text{C}}}}{{0.512{\text{ }}^\circ {\text{C/m}}}}} \right)\\&= 3.05{\text{ m}}\\\end{aligned}$  

Therefore the molality of glucose in the given solution comes out to be 3.05 m.

Learn more:

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Answer details:

Grade: Senior School

Chapter: Colligative properties

Subject: Chemistry

Keywords: colligative properties, boiling point, Kb, 3.05 m, elevation in boiling point, glucose, m, colligative properties, depression in freezing point, osmotic pressure, relative lowering of vapor pressure, temperature change.