For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2ºC. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take Kb = 0.76 K kg mol

Engineering

Chemistry

Raoult Law and Vapour Pressure

Colligative properties and Measurement

Concentration Terms

Question

For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2ºC. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is

(take K_b = 0.76 K kg mol^–1)

736

724

710

718

$∆$ T_b = K_b . m

$or, 2 = 0 .76 \times \frac{2 .5 \times 1000}{M \times 100} \Rightarrow M = \frac{0 .76 \times 25}{2}$

$Now, \frac{P ° - P}{P °} = \frac{n_{solute}}{n_{solvent}} (n_{solvent} > > n_{solute})$

$or, \frac{760 - P}{760} = \frac{(\frac{2 .5}{M})}{(\frac{100}{18})} \Rightarrow P = 724 mm Hg$

This problem involves calculating the vapor pressure of a solution using colligative properties. The key concepts are boiling point elevation and Raoult's law.

Step 1: Find the molar mass of the solute using boiling point elevation

The formula for boiling point elevation is:

Δ T_{b} = K_{b} \times m

where $Δ T_{b}$ is the elevation in boiling point, $K_{b}$ is the ebullioscopic constant, and $m$ is the molality.

Given: $Δ T_{b} = 2$ °C, $K_{b} = 0.76$ K kg mol⁻¹

So, $2 = 0.76 \times m$

Therefore, molality $m = \frac{2}{0.76} = \frac{50}{19}$ mol/kg

Step 2: Relate molality to mole fraction

Molality is defined as moles of solute per kg of solvent. For a dilute solution, mole fraction of solute $x_{2}$ is approximately:

x_{2} = \frac{n_{2}}{n_{1} + n_{2}} \approx \frac{n_{2}}{n_{1}} = m \times M_{1}

where $M_{1}$ is the molar mass of solvent (water) = 0.018 kg/mol.

So, $x_{2} = \frac{50}{19} \times 0.018 = \frac{0.9}{19}$

Step 3: Apply Raoult's law to find vapor pressure

Raoult's law states that the relative lowering of vapor pressure is equal to the mole fraction of solute:

\frac{P_{1}^{0} - P_{s}}{P_{1}^{0}} = x_{2}

where $P_{1}^{0}$ is the vapor pressure of pure solvent and $P_{s}$ is the vapor pressure of solution.

At 1 atm pressure, water boils at 100°C. The vapor pressure of pure water at its boiling point is equal to the atmospheric pressure, which is 760 mm Hg.

So, $P_{1}^{0} = 760$ mm Hg

Therefore,

\frac{760 - P_{s}}{760} = \frac{0.9}{19}

Solving for $P_{s}$ :

760 - P_{s} = 760 \times \frac{0.9}{19} = \frac{684}{19} = 36

Thus, $P_{s} = 760 - 36 = 724$ mm Hg

Final Answer: 724 mm Hg

Detailed Solution

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Detailed Solution

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