Please answer all page. What are the ideal van\'t Hoff factors for the following

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Please answer all page.
What are the ideal van't Hoff factors for the following compounds: C_6H_12O, K_3PO_4, LiNO_3? 1, 1, 1, 1 2, 1, 2, 2 3, 1, 4, 2 6, 3, 5, 5 Consider the following pairs of liquids. Which pairs are miscible? benzene(C_6H_6) and hexane(C_6H_12) water and methanol(CH_3OH) water and hexane I & II I only II only I, II, & III Calculate the freezing point of an aqueous 1.00 m Nacl solution. Assume complete dissociation. (K_r = 1.86 degree C/m) -1.86 degree C 1.86 degree C -3.72 degree C -0.93 degree C A substance that increases the rate of a chemical reaction and may be recovered unchanged at the end of the reaction is a(n) product catalyst activated complex reaction intermediate The reaction A + B rightarrow products is found to be second order in [A] and first order in [B]. The rate equation would be: Rate = k[A][B] Rate = k[A]^2[B] Rale = k[A][B]^2 Rate = k[A]^2 Determine the rate law for the reaction, 2 ICL(g) + H_2(g) rightarrow I_2(g) + 2 HCl(g), from the following initial rate data Rate = k[ICI]^2 Rate = k[H_2]^2 Rate = k[ICI][H_2]^2 Rate = k[ICI][H_2]

Explanation / Answer

6. Vant Hoff factor = number of particles a formula breaks up into. For electrolytes, this simply transforms to i = number of charged particles per molecule of the electrolyte.

Consider the dissociations below:

Ba(OH)2 (s) -----> Ba2+ (aq) + 2 OH- (aq); i = 1 + 2 = 3

C6H12O6 (s) -----> C6H12O6 (aq); i = 1 (non-electrolyte)

K3PO4 (s) -----> 3 K+ (aq) + PO43- (aq); i = 3 + 1 = 4

LiNO3 (s) -----> Li+ (aq) + NO3- (aq); i = 1 + 1 = 2

The correct answer is (c) 3,1,4,2

Ans: (c)

7. (I) Benzene (C6H6) and hexane (C6H12) are miscible since both are non-polar hydrocarbons.

(II) Water (H2O) and methanol (CH3OH) are miscible since both are polar and can form hydrogen bonds.

(III) Water is polar while hexane is non-polar. This pair should be immiscible.

Ans: (A) I and II

8. Let t be the freezing point depression. Given molality of the aqueous solution, m = 1.00 m and the freezing point depression constant, Kf = 1.86C/m, we have,

t = Kf.m = (1.86C/m).(1.00 m) = 1.86C.

Since the normal freezing point of water is 0C and the freezing point depression is 1.86C, we have,

tnormal – tsalt = t

===> 0C – tsalt = 1.86C

===> tsalt = (0 – 1.86)C = -1.86C (ans).

Ans: (A) -1.86C.

9. The substance is called a catalyst. A catalyst can increase the rate of a reaction or decrease the rate; however, the catalyst is recoverable at the end of the reaction.

Ans: (B) catalyst

10. The rate law is Rate = k[A]2[B] where k is the rate constant for the reaction.

Ans: (B) Rate = k[A]2[B]

11. Consider the first and second entries in the table. As [ICl] becomes doubled ([H2] is held constant in the first two entries), the rate of the reaction becomes doubled. Therefore,

Rate [ICl]

Again consider the 3rd and 4th entries where [ICl] is held constant and [H2] is changed. As [H2] becomes doubled in 4 as compared to 3, the rate of reaction 4 is almost twice the rate of reaction 3. Therefore,

Rate [H2]

The rate law is Rate = k[ICl][H2]where k is the second-order rate constant.

Ans: (D) Rate = k[ICl][H2]

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