Question

Calculate the boiling point of a 3.60 m aqueous sucrose solution. express the boiling point in degrees celsius to five significant figures

Answer 1

We can use this equation for boiling point elevation:
ΔT(b) = i K(b) M
when Δ T(b) is the increase of boiling point of the solution.
and i is ( vant Hoff factor, the number of particles or ions per mole-clue.
and K(b) is boiling point increase constant for the solution ( and for water it is equal 0.52 C° Kg/mol)
We can assume i (vant Hoff factor ) = 1 as the sucrose is nonelectrolyte (not readily ionize).
So for water: Tb° = 100 c° and Kb = 0.52 c° Kg / mol
By substitute at:
ΔTb = i Kb M
∴  = 1 * 0.52 * 3.60 = 1.8432 C°
and when Tb = Tb° + ΔTb
∴  Tb = 100 + 1.8432 = 101.8432 C°

Answer 2

Boiling point of 3.60 m aqueous sucrose solution is $\boxed{101.87\text{ }^{\circ}\text{C}}$.

Further Explanation:

Colligative properties are such properties that depend on number of solute particles only, but do not depend on identity of solute. Colligative properties are mentioned below.

1. Relative lowering of vapor pressure

2. Elevation in boiling point

3. Depression in freezing point

4. Osmotic pressure

The formula to calculate 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 elevation in boiling point.

m is the molality of solution.

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

The value of m is 3.60 m.

The value of $\text{k}_{\text{b}}$ is $0.512\text{ }^{\circ}\text{C/m}$.

Substitute 3.60 m for m and $0.512\text{ }^{\circ}\text{C/m}$ for $\text{k}_{\text{b}}$ in equation (1).

$\begin{aligned}\Delta\text{T}_{\text{b}}&=(0.512\text{ }^{\circ}\text{C/m})(\text{3.60 m})\\&=1.872\text{ }^{\circ}\text{C}\end{aligned}$

The boiling point elevation can also be calculated as follows:

$\Delta\text{T}_{\text{b}}=\text{Boiling point of sucrose}-\text{Boiling point of pure water}$      ...... (2)

Rearrange equation (2) for boiling point of sucrose.

$\text{Boiling point of sucrose}=\Delta\text{T}_{\text{b}}+\text{Boiling point of pure water}$      ...... (3)

The value of $\Delta\text{T}_{\text{b}}$ is $1.872\text{ }^{\circ}\text{C}\end{aligned}$.

The value of boiling point of pure water is $100\text{ }^{\circ}\text{C}\end{aligned}$.

Substitute $1.872\text{ }^{\circ}\text{C}\end{aligned}$ for $\Delta\text{T}_{\text{b}}$ and $100\text{ }^{\circ}\text{C}\end{aligned}$ for boiling point of pure water in equation (3).

$\begin{aligned}\text{Boiling point of sucrose}&=1.872\text{ }^{\circ}\text{C}+100\text{ }^{\circ}\text{C}\\&=101.872\text{ }^{\circ}\text{C}\\&\approx{101.87\text{ }^{\circ}\text{C}}\end{aligned}$

Learn more:

1. Which statement is true for Boyle’s law: brainly.com/question/1158880

2. Calculation of volume of gas: brainly.com/question/3636135

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Colligative properties

Keywords: colligative properties, boiling point, osmoctic pressure, depression in freezing point, elevation in boiling point, relative lowering of vapor pressure, solute particles, sucrose, boiling point of sucrose, 3.60 m.