4.0 g molecular PCl5 were heated in a vessel of 6 L volume. At equilibrium state , 50% of PCl5 is dissociated. The equilibrium constant for the reaction is :
0.66 mol L-1
0.33 mol L-1
0.44 mol L-1
0.88 mol L-1
B.
0.33 mol L-1
PCl5 PCl3 + Cl2
Moles before
dissociation C 0 0
Moles after
dissociation C(1 - ) C C
Kc =
Given , C = and = 0.5
Kc =
One mole of FeC2O4 is oxidised by x moles of Cr2 in acidic medium , x is:
0.5
1.0
1.5
2.0
A.
0.5
Cr2+ 14H+ + 6e- 2Cr3+ + 7H2O
FeC2O4 Fe3+ + 2CO2 + 3e- ] x 2
________________________________________
+ + 14H+ 2Cr3+ + 2Fe3+ + 4CO2
+ 7H2O
Thus, one mole of FeC2O4 is oxidised by 0.5 mole of
Cr2 in acidic medium.
The number of bonds in P4O10 is:
6
16
7
17
B.
16
The number of bonds in P4O10 is 16. There are also four bonds presents in P4O10 molecule.
The decreasing nucleophilicity of the following is:
I. CH3S- II. CH3O-
III. -OH IV. EtO-
II > III > 1V > I
III > IV > I > II
IV > III > I > II
II < I < IV < III
C.
IV > III > I > II
Acidity is explained as
CH3SH > CH3OH > H2O > EtOH
Basicity : CH3S- < CH3O- < -OH < EtO-
When nucleophilic centre is different (CH3S- and CH3O-) and they are from the same group, nucleophilicity is opposite to basicity.
Nucleophilicity EtO- > -OH > CH3S- > CH3O-
Hence, the answer is (c)
For a chemical reaction, temperature is raised from 25° C to 35° C. The value of equilibrium constant, Kp is doubled . The heat of reaction (H) for the reaction will be :
13.834 kcal
11.714 kcal
10.726 kcal
4.312 kcal
C.
10.726 kcal
Given that , T1 = 25°C = 298 K
T1 = 35°C = 308 K
= 2R = 2calK-1 mol-1
From van't Hoff equation ,
2.303log =
2.303 log 2 =
or , = 12726 cal = 10.726 kcal
Consider the following pairs of ions:
I. Sc3+ and Ti4+ II. Mn2+ and Fe2+
III. Fe2+ and Co3+ IV. Cu+ and Zn2+
Among these pairs ofions, isoelectronic pairs would include:
II, III and IV
I, III and IV
I, II and IV
I, II and III
B.
I, III and IV
I. Electrons in so3+ = 18 and in Ti4+ = 18
II. Electrons in Mn2+ =23 and in Fe2+ = 24
III. Electrons in Fe2+ = 24 and in Co3+ =24
IV. Electrons in Cu+ = 28 and in Zn2+ = 28
A litre solution containing NH4Cl and NH4OH has hydroxide ion concentration of 10-6 mol L-1. Which of the following hydroxide could be precipitated when the solution is added to 1L solution of 0.1M metal ions?
Ba(OH)2 (Ksp = 5 x 10-3)
Ni(OH)2 (Ksp = 1.6 x 10-16)
Mn(OH)2 (Ksp = 2 x 10-13)
Fe(OH)2 (Ksp = 8 x 10-16)
I and IV
III and IV
II and IV
II , III and IV
C.
II and IV
When 1L each solution are mixed
[OH]- = 10-6 M (buffer solution)
Mn+ = 0.05M
for Q = [0.5][10-6]2 = 5 x 10-14
Q > Ksp for Fe2+ and Ni2+
Thus , Fe(OH)2 and Ni(OH)2 are precipitate.
Arrange the following (I to IV) in the order of increasing masses:
I. 0.5 mole of O3
II. 0.5 g atom of oxygen
III. 3.011 x 1023 mlecules of O2
IV. 5.6 l of CO2 at STP
II < IV < III < I
IV < III < I < II
II < I < IV < III
I < II < III < IV
A.
II < IV < III < I
I. 0.5 mole O3 = 24 g O3
II. 0.5 g atom of oxygen = 8 g O2
III. = 16 g of O2
IV. 44 g CO2 = 11 g CO2
Forty millilitre of CO was mixed 100 mL of O2 and the mixture was exploded. On cooling, the reaction mixture was shaken with KOH. What volume of gas is left?
80 ml of O2
60 ml of O2
80 ml of CO2
50 ml of CO2
A.
80 ml of O2
volume of O2 left = (100-20) = 80 ml
10 moles of ideal gas expand isothermally and reversibly from a pressure of 10 atrn to 1 atm at 300K. What is the largest mass which can lifted through a height of 100 meter?
48.75 kg
58.55 kg
75.64 kg
64.13 kg
B.
58.55 kg
W = - nRT ln
= -10 x 8.314 x 300 ln
= -57441.42 J
W = - mgh
m X 9.81 x 100 = 57441.42
m = 58.55kg
Severity: Notice
Message: Undefined variable: flex_subject_id
Filename: features/flexipad.php
Line Number: 348
Backtrace:
File: /var/www/html/public_html/application/views/features/flexipad.php
Line: 348
Function: _error_handler
File: /var/www/html/public_html/application/views/previous_year_papers/paper.php
Line: 163
Function: view
File: /var/www/html/public_html/application/controllers/Study.php
Line: 2109
Function: view
File: /var/www/html/public_html/index.php
Line: 315
Function: require_once