A 1.35m aqueous solution of compound X had a boiling point of 101.4 degree C. Wh

Question

A 1.35m aqueous solution of compound X had a boiling point of 101.4 degree C. Which one of the following could be compound X? The boiling point elevation constant for water is 0.52 degree C/m. A) CaCI_2 B) Na_3PO_4 C) C_6H_12O_6 D) KCI E) CH_3CH_2OH Which one of the following solutes hat a limiting van't Hoff factor (i) of 3 when dissolved in water? A) CH_3OH B) KNO_3 C) CCI_4 D) sucrose E) Na_2SO_4 At elevated temperatures, dinitrogen pentoxide decomposes to nitrogen dioxide and oxygen 2N_2O_5(g) rightarrow 4NO_2(g) + O_2(g) When the rate of formation of O_2 is 2.2 times 10^-4 M/s, the rate of decomposition of N_2O_5 is M/s. A) 2.8 times 10^-4 B) 5.5 times 10^-4 C) 2.2 times 10^-4 D) 1.1 times 10^-4 E) 4.4 times 10^-4 The data in the table below were obtained for the reaction: The rate law for this reaction is rate =. A) k[A]^2[B]^2 B) k[A]^2 C) k[A][B] D) k[A]^2[B] E) k[P] The reaction CH_3-NaC rightarrow CH_3-CaN Is a first-order reaction. At 230.3 degree C, k = 6.29 times 10^-4 s^-1. If [CH_3-N=C] is 1.00 times 10^-3 initially. [CH_3 - N=C] is after 1.000 times 10^3 s. A) 1.88 times 10^-3 B) 4.27 times 10^-3 C) 1.00 times 10^-6 D) 2.34 times 10^-4 E) 5.33 times 10^-4

Explanation / Answer

(14)

deltaTb =i * Kb * m

101.4 - 100 = i * 0.52 * 1.35

i = Van't Hoff factor = 2, that means the compound should dissociate into two particles.

SO, (D) KCl ---------> K+ + Cl-

(15) (E) Na2SO4 will give three particles on dissociation.

NaSO4 (aq.) --------> 2 Na+ (aq.) + SO42- (aq.)

(16) (E)

r = - (1/2) d[N2O5]/dt = + d[O2]/dt = 2.2* 10-4

d[N2O5]/dt = 4.4 * 10-4 s-1

(17) (B) r = k[A]2

we [B] is doubled no change in r

but when [A] is tripled rate is raised by 9 times.

(18) (E)

k = 2.303/t loga/(a-x)

6.29*10-4*1.000*103 / 2.303 = Log1*10-3 / (a-x)

0.273 = Log10-3/(a-x)

10-3/(a-x) = 1.875

a - x = 5.33 * 10-4 M

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