18 g glucose (C6H12O6) is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution is:

Engineering

Chemistry

Raoult Law and Vapour Pressure

Relative Lowering in Vapour Pressure

Colligative properties and Measurement

Question

18 g glucose (C₆H₁₂O₆) is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution is:

752.4

759.0

76.0

7.6

$n_{C_{6} H_{12} O_{6}} = 0 .1$

$n_{H_{2} O} = \frac{178 .2}{18} = 9 .9$

$\frac{p^{0} - p}{p^{0}} \Rightarrow \frac{0 .1}{10}$

$p^{0} - p \Rightarrow \frac{0 .1}{10} \times 760 = 7 .6$

p = p⁰ – 7.6

= 760 – 7.6

= 725.4 torr

This problem involves calculating the vapor pressure of an aqueous glucose solution. Glucose is a non-volatile solute, so it lowers the vapor pressure of the solvent (water). This phenomenon is described by Raoult's Law.

Raoult's Law for a solution containing a non-volatile solute states that the relative lowering of vapor pressure is equal to the mole fraction of the solute.

The formula is:

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

where:

$P_{1}^{0}$ is the vapor pressure of the pure solvent (water).
$P_{1}$ is the vapor pressure of the solution.
$x_{2}$ is the mole fraction of the solute (glucose).

Step 1: Find the vapor pressure of pure water.
The question does not explicitly state it, but it is a standard value assumed to be 760 torr at 100°C. The large amount of water (178.2 g) is a clue, as it is exactly 10 moles, suggesting the problem is set at a temperature where water's vapor pressure is 760 torr. Therefore, we take $P_{1}^{0} = 760 t o r r$ .

Step 2: Calculate the moles of solute (glucose, C₆H₁₂O₆) and solvent (water, H₂O).
Molar mass of glucose (C₆H₁₂O₆) = (6×12) + (12×1) + (6×16) = 180 g/mol.
Moles of glucose, $n_{2} = \frac{18 g}{180 g / m o l} = 0.1 m o l$ .
Molar mass of water (H₂O) = 18 g/mol.
Moles of water, $n_{1} = \frac{178.2 g}{18 g / m o l} = 9.9 m o l$ .

Step 3: Calculate the mole fraction of the solute (glucose).
Total moles = $n_{1} + n_{2} = 9.9 + 0.1 = 10 m o l$ .
Mole fraction of glucose, $x_{2} = \frac{n_{2}}{n_{1} + n_{2}} = \frac{0.1}{10} = 0.01$ .

Step 4: Apply Raoult's Law to find the vapor pressure of the solution (P₁).
$\frac{760 - P_{1}}{760} = 0.01$
Now, solve for P₁:
$760 - P_{1} = 760 \times 0.01$
$760 - P_{1} = 7.6$
$P_{1} = 760 - 7.6 = 752.4 t o r r$

Final Answer: The vapor pressure of the aqueous solution is 752.4 torr.

Related Concepts & Formulae

Raoult's Law: This law is fundamental for understanding the vapor pressure of ideal solutions. For a solution, the partial vapor pressure of each component is equal to the product of the vapor pressure of the pure component and its mole fraction in the solution. For a solvent with a non-volatile solute, it simplifies to the relative lowering of vapor pressure being equal to the mole fraction of the solute.

Colligative Properties: Properties of solutions that depend on the number of solute particles, not their identity. Lowering of vapor pressure is one such property. Others include boiling point elevation, freezing point depression, and osmotic pressure.

Key Formula:

\frac{ΔP}{P_{1}^{0}} = x_{2} = \frac{n_{2}}{n_{1} + n_{2}}

Detailed Solution

Related Concepts & Formulae

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