Question

The vapor pressure of water at 20 degree Celcius is 17.5mmHg. What is the vapor pressure of water over a solution prepared from 200g of sucrose (${{C}_{12}}{{H}_{22}}{{O}_{11}}$) and 112.3 g of water?

Answer 1

The pressure exerted by a vapour in thermodynamic equilibrium with its condensed phases (solid or liquid) at a particular temperature in a closed system is known as vapour pressure. The evaporation rate of a liquid is determined by the equilibrium vapour pressure. It has to do with particles' proclivity for escaping from liquids (or a solid). Volatile refers to a material that has a high vapour pressure at room temperature. Vapor pressure is the pressure exerted by vapour existing above a liquid surface.

Complete answer:
Raoult's law is a physical chemistry law having thermodynamic consequences. It says that each component in an ideal mixture of liquids has a partial pressure equal to the vapour pressure of the pure component multiplied by its mole fraction in the mixture. As a result, the mole fraction of nonvolatile solute in a dilute solution lowers the vapour pressure compared to the mole fraction of solute in the solution.
Mathematically,
${{p}_{i}}=p_{i}^{\star }{{x}_{i}}$
where ${{p}_{i}}$ is the component i's partial pressure in the gaseous mixture (above the solution), $p_{i}^{\star }$ is the pure component i's equilibrium vapour pressure, and ${{x}_{i}}$ is the component i's mole fraction in the mixture (in the solution).
We utilise Raoult's Law to determine a 16 mm Hg solution vapour pressure.
Raoult's Law states that the vapour pressure of a component in an ideal solution is proportional to the mole fraction of that component in solution.
Mole fraction of the sucrose: $=\dfrac{\text { Moles of sucrose }}{\text { Moles of sucrose }+\text { moles of water }}$
$\chi_{\text {sucrose }}=\dfrac{\dfrac{200 \cdot g}{342.13 \cdot g \cdot m o l^{-1}}}{\dfrac{200 \cdot g}{342.13 \cdot g \cdot m o l^{-1}}+\dfrac{112.3 \cdot g}{18.01 \cdot g \cdot m o l^{l}-1}}=0.0857$
Mole fraction of the water:
$\chi_{\text {water }}=\dfrac{\dfrac{112.3 \cdot g}{18.01 \cdot g \cdot m o l^{-1}}}{\dfrac{200 \cdot g}{342.13 \cdot g \cdot m o l^{-1}}+\dfrac{112.3 \cdot g}{18.01 \cdot g \cdot m o l^{-1}}}=0.914$
By definition, $\chi_{\text {sucrose }}+\chi_{\text {water }}=1$
Because sucrose is inert, its vapour pressure is proportional to the mole fraction of water in the solution:
$P_{\text {solution }}=\chi_{\text {water }} \times 17.5 \cdot m m \cdot H g$
$=0.914 \times 17.5 \cdot m m \cdot H g=16 \cdot m m \cdot H g$
Hence 16mm Hg is the correct answer

Note:
Raoult's law is a phenomenological law that predicts perfect behaviour based on the basic microscopic premise that intermolecular interactions between dissimilar molecules equal those between similar molecules: ideal solution conditions. This is similar to the ideal gas law, which is a limiting law that applies when the intermolecular forces approach zero, such as when the concentration approaches zero. If the physical characteristics of the components are identical, Raoult's law applies. The more related the components are, the more closely their behaviour resembles Raoult's rule.