State Raoult’s law for a solution containing volatile components. Why is the vapour pressure of an aqueous solution of glucose lower than that of water?

Question

$W$ g of a non-volatile electrolyte solid solute of molar mass $M\,\text{g mol}^{-1}$ when dissolved in $100\,\text{mL}$ water decreases vapour pressure of water from $640\,\text{mm Hg}$ to $600\,\text{mm Hg}$. If aqueous solution of the electrolyte boils at $375\,\text{K}$ and $K_b$ for water is $0.52\,\text{K kg mol}^{-1}$, then the mole fraction of the electrolyte $(x_2)$ in the solution can be expressed as ___.
(Given: density of water $=1\,\text{g mL}^{-1}$, boiling point of water $=373\,\text{K}$)

Answer 1

Raoult’s law posits a direct proportionality between the partial vapor pressure of any volatile component within a solution and its corresponding mole fraction. This relationship is expressed mathematically as: \[P_A = X_A \cdot P_A^\circ\]In this equation, $ P_A $ denotes the partial vapor pressure of component A, $ X_A $ represents its mole fraction, and $ P_A^\circ $ signifies the vapor pressure of pure component A.
The vapor pressure of an aqueous glucose solution is diminished compared to that of pure water due to glucose being a non-volatile solute. An increase in glucose molecules within the solution leads to a reduction in the number of water molecules at the surface, consequently lowering the vapor pressure. This phenomenon illustrates a colligative property, where the depression in vapor pressure is directly proportional to the quantity of solute present.

State Raoult’s law for a solution containing volatile components. Why is the vapour pressure of an aqueous solution of glucose lower than that of water?

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Solution and Explanation

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