The freezing point of benzene decreases by 0.45

The freezing point of benzene decreases by 0.45\(^0\)C when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be. (K\(_f\) for benzene=5.12 K kg mol\(^\)) - Find Answer & Solution | LearnPick Resources Disclaimer: Students can find the best tutors and instructors through LearnPick's online tutoring marketplace. We neither supply nor recommend tutors to those in search of such services, and vice-versa. LearnPick does not verify the identity or authenticity of information posted by tutors or students. You can learn more about verifying the identity of other users in our Safety Center.

The freezing point of benzene decreases by $$ ]A.$74.6\\% $ B.$94.6\\% $ C.$64.6\\% $ D.$80.4\\% $ The freezing point of benzene decreases by $$ ] A.$74.6\% $ B.$94.6\% $ C.$64.6\% $ D.$80.4\% $ Hint: When a non - volatile solute is added to a volatile solvent , then vapour pressure decreases , boiling point increases and freezing point decreases . Freezing point of a substance is a colligative property . Formula used : $\Delta $ depends upon the nature of the solvent , that is , different solvents have different values of molal depression constant and its value does not change for a given solute.

Question Updated on: 13/06/2023 When 0.4 g of acetic acid is dissolved in 40 g of benzene, the freezing point of the solution is lowered by 0.45 K. Calculate the degree of association of acetic acid. Acetic acid forms dimer when dissolved in benzene. ( K f for benzene = 5.12 K kg m o l − 1 at.wt. C = 12, H = 1, O = 16) Molecular mass (normal) for acetic acid, C H 3 C O O H is, M normal = 60 g mol − 1 Observed molecular mass of C H 3 C O O H. M observed = K f w 2 w 1 Δ T f = ( 5.12 K kg mol − 1 ) ( 0.4 × 10 − 3 k g ) ( 0.04 k g ) ( 0.45 K ) = 113.77 × 10 − 3 kg mol = 113.77 g mol − 1 Now, van.t Hoff factor, i = M normal M observed = 60 g mol − 1 113.77 g mol − 1 ⇒ 0.527 Let x be the degree of association of C H 3 C O O H, then, 2 C 6 H 5 C O O H ↔ ( C 6 H 5 C O O H ) 2 If x represents the degree of association of the solute then we would have (1 - x) mol of benzoic acid left in unassociated form and correspondingly x/2 as associated moles of benzoic acid at equilibrium. Therefore, The total number of moles after association = 1 − x + x 2 = 1 − x 2 Normal molar mass Observed molar mass = 1 − x 2 60 g mol − 1 113.77 g mol − 1 = 1 − x 2 0.527 = 1 − x 2 x 2 = 1 − 0.527 x = 0.473 × 2 = 0.946 = 0.95 Hence, degree of association, x=0.95. or degree of association of acetic acid is 95%. ab question is when 0.4 G of Acetic acid is dissolved in 40 G of benzene the freezing point of the solution is lowered by 0.45 Kelvin calculate the degree of dissociation of Acetic Acid Acetic Acid f...

The correct option is D 94.6% Let the degree of association of acetic acid ( C H 3 C O O H ) in benzene is α, then 2 C H 3 C O O H ⇌ ( C H 3 C O O H ) 2 I n i t i a l m o l e s 1 0 M o l e s a t e q u i l i b r i u m 1 − α α 2 ∴ T o t a l m o l e s = 1 − α + α 2 = 1 − α 2 o r i = 1 − α 2 Now, depression in freezing point ( Δ T f ) is given as Δ T f = i K f m . . . . (i) Where, K f - molal depression constant or cryoscopic constant. m = Molality M o l a l i t y = n u m b e r o f m o l e s o f s o l u t e w e i g h t o f s o l v e n t ( i n k g ) = 0.2 60 × 100 20 Putting the values in Eq.(i) ∴ 0.45 = [ 1 − α 2 ] ( 5.12 ) [ 1000 20 ] 1 − α 2 = 0.45 × 60 × 20 5.12 × 0.2 × 1000 ⇒ 1 − α 2 = 0.527 ⇒ α 2 = 1 − 0.527 ∴ α = 0.946 Thus, percentage of association = 94.6%

Your strategy here will be to • determine the van't Hoff factor for sodium iodide, #"NaI"# • calculate the molality of the solution • calculate the freezing-point depression of the solution The idea is that the freezing point of a solution is lower than the freezing point of the pure #0^@"C"# at normal pressure. Now, the difference between the freezing point of the pure solvent and the freezing point of the solution is given by the freezing-point depression, which can be calculated using the equation #color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = i * K_f * bcolor(white)(a/a)|)))# Here #DeltaT_f# - the freezing-point depression; #i# - the van't Hoff factor #K_f# - the cryoscopic constant of the #b# - the The problem provides you with the cryoscopic constant for water #K_f = 1.86^@"C kg mol"^(-1)# Now, sodium iodide is soluble in aqueous solution, which means that it dissociates completely to form sodium cations, #"Na"^(+)#, and iodide anions, #"I"^(-)# #"NaI"_ ((aq)) -> "Na"_ ((aq))^(+) + "I"_ ((aq))^(-)# Notice that every mole of sodium iodide that is dissolved in solution produces two moles of particles of This means that the van't Hoff factor, which tells you the ratio that exists between how many moles of solute you're dissolving and the number of moles of particles of solute that are produced in solution, will be equal to #2# #i = 2 -># one mole of solute dissolved, two moles of ions produced The one kilogram of solvent. Your solution contains #0.5# moles of sodium i...

The correct Answer is B Δ T f (observed) = 0.45 ∘ Δ T f (calculated) = 1000 K f w 2 w 1 × M 2 = 1000 × 5.12 × 0.2 20 × 60 ( M 2 ( C H 3 C O O H ) = 60 ) = 0.853 ∘ ∴ = Δ T f (observed) Δ T f (calculated) = 0.45 0.853 = 0.527 2 C H 3 C O O H 1 − α ⇔ ( C H 3 C O O H ) 2 α 2 Total moles = 1 − α + α 2 = 1 − α 2 i = 1 − α 2 = 0.527 or α 2 = 0.473 or α = 0.946 % association = 94.6 %