At a certain temperature the following equilibrium is established, CO(g) + N0₂(g) $\stackrel{}{\rightleftharpoons }$ CO₂(g)+ NO(g)One mole of each of the four gas is mixed in one litre container and the reaction is allowed to reach equilibrium state. When excess of baryta water (Ba(OH)₂) is added to the equilibrium mixture, the weight of white ppt (BaCO₃) obtained is 236.4 gm. The equilibrium constant K_c of the reaction is (Ba = 137)

(1) 1.2

(2) 2.25

(3) 2.1

(4) 3.6

The pOH of a solution at 25$°$C that contains 1 x 10^-10 M of hydronium ions is:  

1. 7.00                               

2. 4.00

3. 9.00                                 

4. 1.00

If the concentration of OH^- ions in the reaction

Fe(OH)³ (s) $\rightleftharpoons$Fe³⁺ (aq) + 3OH^-(aq) is decreased to 1/4 times, then equilibrium concentration of Fe³⁺ will increase by

(a) 8 times                            (b) 16 times

(c) 64 times                          (d) 4 times

Which one of the following molecular hydrides acts as a Lewis acid?

1. NH₃         

2. H₂O         

3. B₂H₆           

4. CH₄

K_c and K_p are not equal for a reaction given below:

1. 2NO(g) $\rightleftharpoons$N₂(g) + O₂(g)

2. SO₂(g) + NO₂(g) $\rightleftharpoons$SO₃(g) + NO(g)

3. H₂(g) + I₂(g) $\rightleftharpoons$2HI(g)

4. 2C(s) + O₂(g) $\rightleftharpoons$2CO₂(g)

The molecule that is least likely to behave as Lewis base is: 

1. NH₃

2. BF₃

3. OH^-

4. H₂O

If carbon dioxide is 2% dissociated at equilibrium 2CO₂ (g)$\rightleftharpoons$2CO(g) + O₂(g). The mole fraction of CO₂ at equuilibrium is

1. 1.01/0.98           

2. 0.98/1.01

3. 0.01/0.98           

4. 0.098/1.01

Ostwald dilution law for weak electrolyte HA can be given as

$\left(1\right)<mspace\; width="1em"></mspace>{\mathrm{K}}_{\mathrm{a}}=\frac{{\mathrm{C\alpha }}^2}{(1-\mathrm{\alpha })}$
$\left(2\right)<mspace\; width="1em"></mspace>\mathrm{Ka}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>{\mathrm{C\alpha }}^2<mspace\; width="1em"></mspace>\mathrm{if}<mspace\; width="1em"></mspace>\mathrm{\alpha }\to 0$
$\left(3\right)<mspace\; width="1em"></mspace>{\mathrm{K}}_{\mathrm{a}}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\frac{{\mathrm{C\lambda }}^2}{{\mathrm{\lambda }}^2({\mathrm{\lambda }}^2-\mathrm{\lambda })}$
$\left(4\right)<mspace\; width="1em"></mspace>\mathrm{All}<mspace\; width="1em"></mspace>\mathrm{of}<mspace\; width="1em"></mspace>\mathrm{the}<mspace\; width="1em"></mspace>\mathrm{above}$

The amount of CaC₂O₄ (molecular weight =128) will dissolve in distilled water to make 1 liter of a saturated solution is: 

[K_sp (CaC₂O₄) = 2.5 x 10^-9 ]

1. 0.0064 g             
2. 0.1280 g
3. 0.0128 g           
4. 1.2800 g

$\frac{{\mathrm{K}}_{\mathrm{p}}}{{\mathrm{K}}_{\mathrm{c}}}$ for the gaseous reaction:

$\mathrm{I}.<mspace\; width="1em"></mspace>2\mathrm{A}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>3\mathrm{B}<mspace\; width="1em"></mspace>\rightleftharpoons 2\mathrm{C}$
$\mathrm{II}.<mspace\; width="1em"></mspace>2\mathrm{A}\rightleftharpoons 4\mathrm{B}$
$\mathrm{III}.<mspace\; width="1em"></mspace>\mathrm{A}+\mathrm{B}+2\mathrm{C}<mspace\; width="1em"></mspace>\rightleftharpoons 4\mathrm{D}$

Would be respectively:

1. (RT)^-3, (RT)², (RT)⁰ 

2. (RT)^-3, (RT)-², (RT)^-1 

3. (RT)^-3, (RT)², (RT) 

4. None of the above