What type of deviation is shown by a mixture of ethanol and acetone

(a) Positive deviation is shown by a mixture of ethanol and acetone. In pure ethanol, a very high fraction of molecules are hydrogen bonded. On adding acetone, its molecules, get in between the molecules of ethanol, thus breaking the hydrogen bonds and reducing ethanol-ethanol attractions considerably. As a result escaping tendency of molecules into the vapour phase increases. Partial vapour pressure of each component and total vapour pressure of the solution comes out to be more than that expected from Raoult's law. Hence, this solution shows positive deviation. … H − O | C 2 H 5 … H − O | C 2 H 5 … H − O | C 2 H 5 … (b) Mass of glucose = 10 g Mass of solution = 100 g Mass of solvent = 100 - 10 = 90 g Mol. mass of glucose = 18 g m o l − 1 Vol. of solution = M a s s D e n s i t y = 100 1.2 = 1000 12 = 83.33 c m 3 Molarity = 10 180 × 1000 83.3 = 0.67 M Molality = 10 180 × 1000 90 = 0.62 m

So, my teacher said, "For a solution to show positive deviation from Raoult's Law, it must have a compound which is lacking hydrogen bond and a compound which has hydrogen bond." This is because the compound not having a hydrogen bond breaks the hydrogen bond of the other compound. Now, why does ethanol and water mixture show positive deviation? Both of them have hydrogen bonds. In more generality, Raoult's law is stated as $$P_i = x_iP_\textb$ interactions. You should see now that your teacher's statement is a general rule-of-thumb following from the concepts outlined above, but does not necessarily hold true for all situations. Here, water forms hydrogen bonds of greater strength and magnitude than does methanol, and one expects that a water-methanol mixture will therefore possess weaker intermolecular interactions than those in a solution of pure water.

4. (i) What type of deviation is shown by a mixture of ethanol and acetone? Give reason. (ii) A solution of glucose (molax mass = 180 g mol − 1 ) in water is labelled as 10% (by mass). What would be the molality and molarity of (Density of solution = 1.2 g mL − 1 ) the solution? Ans. (i) Mixture of ethanol and acetone shows positive deviations from Raoult's law. This is because in ethanol, the molecules are held together due to hydrogen bonding as: When acetone is added to ethanol, there are weaker interactions between acetone and ethanol than ethanol-ethanol interactions. Some molecules of acetone occupy spaces between ethanol molecules and consequently, some hydrogen bonds in alcohol molecules break and attractive forces between ethanol molecules are weakened. Therefore, the escaping tendency of ethanol and acetone molecules from solution increases. Thus, the vapour pressure of the solution is greater than the vapour pressure as expected according to Raoult's law. (ii) W B ​ = 10 g, wt. of solvent = 90 g , M B ​ = 180 g mol − 1 Molality ​ = Wt. of solvent Moles of solute ​ × 1000 = 180 × 90 10 × 1000 ​ = 0.617 m ​ Volume of solution = 1.2 100 ​ = 83.3 mL Molarity ​ = Vol. of solution (in mL ) Moles of solute ​ × 1000 = 180 10 ​ × 83.3 1000 ​ = 0.66 M ​ ​ 5. (a) A 5% solution (by mass) of cane sugar in water has a freezing point of 271 K. Views: 5,559 MeOH It ionises in water as follows: MCOH ⇌ Me + + OH − Yerrow Red In presence of acich, Or −ions giniu by rethylorange ar...