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Colligative Properties I

Student can calculate molar mass using Relative Lowering of Vapour Pressure and Elevation of Boiling Point.

7 lessons

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Colligative Properties & RLVP

Defines colligative properties and Relative Lowering of Vapour Pressure.

Colligative Properties

Many properties of solutions depend only on the number of solute particles irrespective of their nature relative to the total number of particles present in the solution.

Such properties are called colligative properties (from Latin: co means together, ligare means to bind). Examples include the lowering of vapour pressure, elevation of boiling point, and depression of freezing point.

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Example 1.6: Molar Mass from RLVP

Calculates molar mass of a solid from vapour pressure data.

Problem

The vapour pressure of pure benzene at a certain temperature is $0.850\text{ bar}$ . A non-volatile, non-electrolyte solid weighing $0.5\text{ g}$ when added to $39.0\text{ g}$ of benzene (molar mass $78\text{ g mol}^{-1}$ ). Vapour pressure of the solution, then, is $0.845\text{ bar}$ . What is the molar mass of the solid substance?

Given:

$p_1^0 = 0.850\text{ bar}$ (Pure solvent vapour pressure)
$p_1 = 0.845\text{ bar}$ (Solution vapour pressure)
$w_2 = 0.5\text{ g}$
$w_1 = 39.0\text{ g}$
$M_1 = 78\text{ g mol}^{-1}$

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Practice: RLVP

Solves for vapour pressure lowering.

Adaptation of Intext 1.9: 50 g of urea (

M_2 = 60\text{ g/mol}

) is dissolved in 850 g of water (

M_1 = 18\text{ g/mol}

). The vapour pressure of pure water at 298 K is 23.8 mm Hg. Calculate the relative lowering of vapour pressure and the vapour pressure of the solution.

Given Data

List all the known values with their units.

Formula

State the formula for Relative Lowering of Vapour Pressure.

Substitution

Substitute the values into your formula.

Calculation

Show your step-by-step math to find both answers.

Final Answer

Provide the final numerical answers with units.

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Vapour Pressure Curve (Boiling Point)

Graph showing boiling point elevation.

A clean scientific coordinate-graph diagram showing vapour pressure versus temperature. The y-axis represents Vapour pressure and the x-axis represents Temperature in Kelvin. Two curves are plotted: the upper curve is for pure solvent, and the lower curve is for the solution. A horizontal dashed line marks 1.013 bar (1 atm) pressure. Vertical dashed lines drop from the intersection of the curves with the 1 atm line to the x-axis, marking T_b^0 for the pure solvent and T_b for the solution. The difference between these temperatures on the x-axis is labeled as delta T_b, visually proving the elevation of boiling point.

Click to zoom

The graph shows elevation of boiling point.

A liquid boils when its vapour pressure becomes equal to atmospheric pressure, about $1.013\ \text{bar}$ or $1\ \text{atm}$ .

The solution curve lies below the pure solvent curve, so the solution has lower vapour pressure at the same temperature. Therefore, it needs a higher temperature to boil.

The boiling point of pure solvent is $T_b^0$ , while that of the solution is $T_b$ .

\Delta T_b = T_b - T_b^0

Here, $\Delta T_b$ is the elevation of boiling point.

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Elevation of Boiling Point

Explains boiling point elevation using the molal elevation constant.

The boiling point of a liquid is the temperature at which its vapour pressure equals the atmospheric pressure.

Because adding a non-volatile solute lowers the vapour pressure of a solvent, you must heat the solution to a higher temperature to make it boil. This increase is called the elevation of boiling point ( $\Delta T_b$ ).

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Example 1.7: Boiling Point Elevation

Calculates the new boiling point of water with glucose.

Problem

$18\text{ g}$ of glucose, $C_6H_{12}O_6$ , is dissolved in $1\text{ kg}$ of water in a saucepan. At what temperature will water boil at $1.013\text{ bar}$ ? $K_b$ for water is $0.52\text{ K kg mol}^{-1}$ .

Given:

$w_2 = 18\text{ g}$
$M_2 = 180\text{ g mol}^{-1}$
$w_1 = 1\text{ kg} = 1000\text{ g}$
$K_b = 0.52\text{ K kg mol}^{-1}$

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Worksheet

Molar Mass from BP

Calculates molar mass from boiling point elevation data.

0 of 3 blanks filled

The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute is dissolved in 90 g of benzene, the boiling point is raised to 354.11 K.

We are given $K_b$ for benzene as $2.53 \text{ K kg mol}^{-1}$ . First, calculate the elevation in boiling point ( $\Delta T_b$ ). $\Delta T_b = 354.11 \text{ K} - 353.23 \text{ K} =$ K.

Next, substitute these values into the molar mass expression $M_2 = \frac{K_b \times w_2 \times 1000}{\Delta T_b \times w_1}$ . $M_2 = \frac{2.53 \times 1.80 \times 1000}{ ␟blank2␟ \times 90 }$ .

Finally, calculate the molar mass of the solute, yielding $M_2 =$ $\text{g mol}^{-1}$ .

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Example 1.6: Molar Mass from RLVP

Calculates molar mass of a solid from vapour pressure data.

Problem

Given:

$p_1^0 = 0.850\text{ bar}$ (Pure solvent vapour pressure)
$p_1 = 0.845\text{ bar}$ (Solution vapour pressure)
$w_2 = 0.5\text{ g}$
$w_1 = 39.0\text{ g}$
$M_1 = 78\text{ g mol}^{-1}$

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feedback

Practice: RLVP

Solves for vapour pressure lowering.

Adaptation of Intext 1.9: 50 g of urea (

M_2 = 60\text{ g/mol}

) is dissolved in 850 g of water (

M_1 = 18\text{ g/mol}

). The vapour pressure of pure water at 298 K is 23.8 mm Hg. Calculate the relative lowering of vapour pressure and the vapour pressure of the solution.

Given Data

List all the known values with their units.

Formula

State the formula for Relative Lowering of Vapour Pressure.

Substitution

Substitute the values into your formula.

Calculation

Show your step-by-step math to find both answers.

Final Answer

Provide the final numerical answers with units.

IMAGE

Vapour Pressure Curve (Boiling Point)

Graph showing boiling point elevation.

Click to zoom

The graph shows elevation of boiling point.

A liquid boils when its vapour pressure becomes equal to atmospheric pressure, about $1.013\ \text{bar}$ or $1\ \text{atm}$ .

The solution curve lies below the pure solvent curve, so the solution has lower vapour pressure at the same temperature. Therefore, it needs a higher temperature to boil.

The boiling point of pure solvent is $T_b^0$ , while that of the solution is $T_b$ .

\Delta T_b = T_b - T_b^0

Here, $\Delta T_b$ is the elevation of boiling point.

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Elevation of Boiling Point

Explains boiling point elevation using the molal elevation constant.

The boiling point of a liquid is the temperature at which its vapour pressure equals the atmospheric pressure.

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Example 1.7: Boiling Point Elevation

Calculates the new boiling point of water with glucose.

Problem

Given:

$w_2 = 18\text{ g}$
$M_2 = 180\text{ g mol}^{-1}$
$w_1 = 1\text{ kg} = 1000\text{ g}$
$K_b = 0.52\text{ K kg mol}^{-1}$

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Worksheet

Molar Mass from BP

Calculates molar mass from boiling point elevation data.

0 of 3 blanks filled

The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute is dissolved in 90 g of benzene, the boiling point is raised to 354.11 K.

We are given $K_b$ for benzene as $2.53 \text{ K kg mol}^{-1}$ . First, calculate the elevation in boiling point ( $\Delta T_b$ ). $\Delta T_b = 354.11 \text{ K} - 353.23 \text{ K} =$ K.

Next, substitute these values into the molar mass expression $M_2 = \frac{K_b \times w_2 \times 1000}{\Delta T_b \times w_1}$ . $M_2 = \frac{2.53 \times 1.80 \times 1000}{ ␟blank2␟ \times 90 }$ .

Finally, calculate the molar mass of the solute, yielding $M_2 =$ $\text{g mol}^{-1}$ .