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Interpretation: The reason for the low vapor pressure of solution than that of pure solvent is need to be defined and described.
Concept introduction: Colligative properties are the properties of a solution that depend upon the number of solutes and not on the nature of the solute. The three colligative properties of the solution are lowering in vapor pressure, depression in freezing point, and elevation in boiling point.
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Answer to Problem 29LC
A solution has a lower vapor pressure than a solution due to less number of solute particles in the vapor state than that of a pure solvent.
Explanation of Solution
Lowering vapor pressure is one of the colligative properties of the solution. This property is described as follows:
Lowering of vapor pressure: Vapor pressure is defined as the pressure exerted by the vapor on its liquid in a closed system. In a closed system, the vapors remain in dynamic equilibrium with their liquid. When the non-volatile solute is added to the solvent then the vapor pressure of the solution comes out to be less than that of the pure solvent. This is because a non-volatile solute remains in its liquid state and very fewer molecules go to the vapor state hence, the pressure exerted by the vapors becomes less and the vapor pressure of the solution comes out to be less than that of the pure solvent.
For example: If glucose or any other non-volatile solute like sodium chloride is added to the solvent then, the vapor pressure of the solution decreases. This is because glucose and sodium chloride both are non-volatile solutes.
A solution has a lower vapor pressure than that of a solution due to less number of solute particles in a vapor state than that of the pure solvent.
Chapter 16 Solutions
Chemistry 2012 Student Edition (hard Cover) Grade 11
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- I have a excitation/emission spectra of a quinine standard solution here, and I'm having trouble interpreting it. the red line is emission the blue line is excitation. i'm having trouble interpreting properly. just want to know if there is any evidence of raman or rayleigh peaks in the spectra.arrow_forwardGive the major product of the following reaction. excess 1. OH, H₂O 1.OH H CH3CH2CH21 H 2. A.-H₂O Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced Template toolbars. The single bond is active by default.arrow_forward2. Use Hess's law to calculate the AH (in kJ) for: rxn CIF(g) + F2(g) → CIF 3 (1) using the following information: 2CIF(g) + O2(g) → Cl₂O(g) + OF 2(g) AH = 167.5 kJ ΔΗ 2F2 (g) + O2(g) → 2 OF 2(g) 2C1F3 (1) + 202(g) → Cl₂O(g) + 3 OF 2(g) о = = -43.5 kJ AH = 394.1kJarrow_forward
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