Class-12-science»Chemistry
On adding cyclohexane to ethanol . Its molecules get in between the molecules of ethanol thus breaking the hydrogen bonds and reducing ethanol - ethanol interaction . Thus will increase the vapour pressure of the solution and result in positive deviation from Rouault's law.
Answer
Hint: These liquid pairs show positive deviation from Raoult's Law because when we add cyclohexane to the ethanol it will break the hydrogen bonding between the ethanol molecules and will reduce the ethanol-ethanol interaction. Vapor pressure increases in this case and heat energy will be absorbed.
Complete step by step solution:
From your chemistry lessons you have learned about the Raoult's law, non- ideal gases showing positive and negative deviations from the Raoult's law.According to Raoult's law partial pressure of a particular component is directly proportional to the mole fraction of that component in liquid solution and is given as:\[{{P}_{A}}=P_{A}^{{}^\circ }{{X}_{A}}\]\[{{P}_{B}}=P_{B}^{{}^\circ }{{X}_{B}}\]-Where ${{P}_{A\,}}\,and\,{{P}_{B}}$= partial vapor pressure of A and B,$P_{A}^{{}^\circ }\,and\,P_{B}^{{}^\circ }$= pure vapor pressure of A and B, and ${{X}_{A}}\,and\,{{X}_{B}}$= mole fraction-In this question we will deal with the non- ideal gas showing positive deviation from Raoult's law. Non- ideal solutions are those which do not obey Raoult's law. Positive deviation from Raoult's law takes place when the vapor pressure of the components will be greater than the expected value in Raoult's law i.e.${{P}_{A}}>P_{A}^{{}^\circ }{{X}_{A}}\,and\,{{P}_{B}}>\,P_{B}^{{}^\circ }{{X}_{B}}$.-In positive deviation the solute-solvent force of attraction will be weaker than the solute-solute interaction and solvent-solvent interaction. A – B < A -A or B - B-Therefore, when we add the cyclohexane in ethanol the molecules of cyclohexane which comes in between the molecules of ethanol will break the hydrogen bonds and thus the interaction between the ethanol-ethanol molecule will get reduced.-In positive deviation the enthalpy of mixing will be positive i.e.$\Delta {{H}_{mix}}>0$. Here the heat will get absorbed and have a positive sign (+).Thus the correct option will be (A).
Note: Ideal solutions are those solutions which obey Raoult's law at all temperatures and at every range of concentration. In this type of solution the force of attraction between the solute-solvent and solute-solute and solvent-solvent is equal. Here the enthalpy of mixing will be zero. Examples of Ideal solutions are n-hexane and n-heptane, Benzene and Toluene.
Vapour pressure of pure water at 298 K is 23.8 mm. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
Vapoure pressure of pure water (solvent) at 298 K, p0 = 23.8 mmVapour pressure of solution, p = ?Mass of solvent ,W = 850 gMass of solute,M = 50 g
Mol. mass of water (H2O), M = 18 g mol–1
Mol.mass of urea NH2 CO NH2
= 14 + 2 + 12 + 16 + 14 + 2
= 60 g mol–1
According to Raoult's law, p0-pp0=ωMWm
p=p0-w×Mm×W×p°
p=23.8-50×1860×850
=23.8-0.017=23.78
Hence, 23.78 mm Hg. Ans.