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Question
CBSEENCH11008059
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The correct set of four quantum numbers for the valence electrons fo rubidium atom (Z = 37) is

  • 5,0,0, +1/2

  • 5,1,0,6+1/2

  • 5,1,1,+1/2

  • 5,0,1,+1/2

Solution

A.

5,0,0, +1/2

the atomic number of Rb (Z) = 37
Thus, its electronic configuration [Kr]5s1. since the last electron or valence electron enters in 5s subshell. So, the quantum numbers are n = 5, l =0(for s orbital) m=0 
(m = +l to -l), s = +1/2 or -1/2

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Question
CBSEENCH11008060
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If Z is a compressibility factor, Vander Waal's equation at low pressure can be written as 

  • straight Z space equals 1 plus RT over pb
  • straight Z space equals space 1 minus straight a over VRT
  • straight Z space equals space 1 minus pb over RT
  • straight Z space equals space 1 space plus space pb over RT

Solution

B.

straight Z space equals space 1 minus straight a over VRT

According to Vander Waal's equation
open parentheses straight p plus straight a over straight V squared close parentheses space left parenthesis straight V minus straight b right parenthesis space equals space RT At space low space pressure comma space open parentheses straight p plus straight a over straight V squared close parentheses straight V space equals space RT pV space plus straight a over straight V space equals space RT pV space equals space RT minus straight a over straight V Divide space both space side space by space RT comma pV over RT space equals space 1 minus straight a over RTV

Question
CBSEENCH11008061
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For the complete combustion of ethanol, C2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2O (l), the amount of heat produced as measured in a bomb calorimeter, is 1364.47 kJ mol-1 at 25oC. Assuming ideality the enthalpy of combustion, ∆CH, for the reaction will be (R = 8.314 JK-1 mol-1)

  • -1366.95 kJ mol-1

  • -1361.95 kJ mol-1

  • -1460.50 kJ mol-1

  • -1350.50 kJ mol-1

Solution

A.

-1366.95 kJ mol-1

C2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2O (l),
∆U = - 1364.47 kJ/mol
∆H = ∆U +∆ngRT
∆ng = -1
∆H = - 1364.47 +fraction numerator negative 1 space straight x space 8.314 space straight x space 298 over denominator 1000 end fraction
[Here, value of R in unit of J must be converted into kJ]
 = - 1364.47-2.4776
 = -1366.94 kJ/mol

Question
CBSEENCH11008062
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For the reaction,

SO subscript 2 space left parenthesis straight g right parenthesis space plus space 1 half space straight O subscript 2 space left parenthesis straight g right parenthesis space leftwards harpoon over rightwards harpoon space SO subscript 3 space left parenthesis straight g right parenthesis
if Kp = Kc (RT)x where the symbol has usual meaning then the value of x is (assuming ideality)

  • -1

  • -1/2

  • 1/2

  • 1

Solution

B.

-1/2

For  the given reaction, ∆ng = np-nR
where np = number of moles of products
nR = number of moles of reactants
Kp = Kc (RT)∆ng
∆ng = -1/2

Question
CBSEENCH11008063
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In which of the following reactions H2O2 acts as a reducing agent?
I. H2O2 + 2H+ + 2e- →2H2O
II. H2O2 - 2e- →O2 + 2H+
III. H2O2 +2e-→ 2OH-
IV. H2O2+ 2OH- -2e- →O2 + 2H2O

  • I and II

  • III and IV

  • I and III

  • II and IV

Solution

D.

II and IV

The release of the electron is known as reduction. So, H2O2 acts as a reducing agent when it releases electrons.
Here, in reactions (II) and (IV), H2O2 release two electrons, hence reactions (II)  and (IV) is known as reduction. In reactions (I) and (III), two electrons are being added so (I) and (III) represents oxidation.

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