Spontaneous adsorption of gas on solid surface is an exothermic process because :
H increases for system
S increases for gas
S decreases for gas
G increases for gas
C.
S decreases for gas
When a gas is adsorbed spontaneously on a solid surface, entropy decreases for gas.
1 mol of O2 and x mol of Ne in a 10 L flask at constant temperature exert a pressure of 10 atm. If partial pressure of O2 is 2 atm , moles of Ne in the mixture is :
1
2
4
3
C.
4
D.
3
pO2 = ptotal .
=
= 2 = 10 x or 10 = 2 + 2x
x = 8/2 = 4 mol
0.1 M solution of CH3COOH should be diluted to how many times so that pH is doubled ?
4.0 times
5.55 x 104 times
5.55 x 106 times
10-2 times
B.
5.55 x 104 times
pH = [pKa - log1] =
pH' (twice of pH) = pKa
pKa = [pKa - logC]
logC = pKa = -logKa
C = Ka = 1.8 x 10-5 M
Dilution = = = 5.55 x 104 times.
The heat of formation of C12H22O11 (s) , CO2 (g) and (H2O) are - 530 , -94.3 and- 68.3 kcalmol-1 respectively. The amount of C12H22O11 to supply 2700 kcal of energy is :
382.70 g
832.74 g
463.9 g
684.0 g
D.
684.0 g
C12H22O11 (s) + 12O2 (g) 12CO2 (g) + 11H2O (l)
= [12(CO2) + 11 (H2O)] - (C12H22O11)]
= - 1352.9 kcal
Thus, number of moles of C12H22O11 for getting 2700 kcal of heat.
= = 2 mol
= 2 X 242 = 684.0 g
NO2 (brown colour gas) exists in equilibrium with N2O4 (colourless gas) as given by chemical equation.
Mixture is slightly brown due to existence of NO2. If pressure is increased :
colour intensity is increased
colour intensity is decreased
colour intensity first increases and then decreases
No change in colour intensity
D.
No change in colour intensity
By Le-Chateliers principle, due to increase in pressure, equilibrium shifts in the direction of low pressure (in which number of mole decreases) i.e. , forward side.
If the equilibrium constant of the reaction of weak acid HA with strong base is 109 then pH of 0.l M NaA is :
5
9
7
8
B.
9
HA + OH- H2O + A-
K =
HA H+ + A-
Ka =
= [H+][OH-] = Kw
or Ka = K2K = 10-14 109 = 10-5
pKa = 5
A- solution is alkaline due to hydrolysis
pH = 7 + +
= 7 + + = 9
For the process ,
X (g) + e- X- (g) ; H = x
and X- (g) X (g) + e- ; H = y
select correct alternate
Ionisation energy of x- (g) is y
Electron affinity of X (g) is x
Electron affinity of X (g) is - y
All of the above
D.
All of the above
X (g) + e- X- (g) ; H = x = EA of X (g)
X- (g) X (g) + e- ; H = y = IE of X- (g)
= - EA of X (g)
Therefore , all options are correct .
A 1 L flask contains 32 g O2 gas at 27C. What mass of O2 must be released to reduce the pressure in the flask to 12.315 atm?
8 g
16 g
24 g
32 g
B.
16 g
pV = nRT = RT
w = = = 16 g
O2 to be released= 32 - 16 = 16 g
Mass of one atom is 6.66 x 10-23 g. Its percentage in an hydride is 95.24. Thus , hydride is :
MH
MH2
MH3
MH4
B.
MH2
Mass of one atom= 6.66 x 10-23 g
Mass of N0 atoms = 6.66 x 10-23 x 6.02 x 10-23 g = 40 g
Thus , atomic weight of the element = 40
Element | % | % at.wt. | Ratio |
M | 95.24 | 2.381 | 1 |
H | 4.76 | 4.76 | 2 |
Thus , hydride is MH2 .
(C2H4) , (C2H6) are x1 and x2 kcalmol-1 respectively. Then heat of hydrogenation of C2H4 is :
x1 + x2
x1 - x2
-(x1 - x2)
x1 + 2x2
C.
-(x1 - x2)
C2H4 + H2 C2H6
H = (C2H6) - (C2H4) - (H2)
= x2 - x1 - 0
= x2 - x1
= -(x1 - x2)