Lakshya Education MCQs

Question: The vapor density of completely dissociated NH4Cl would be
Options:
A.same as that of NH4Cl
B.double than that of NH4Cl
C.half than that of NH4Cl
D.slightly less than that of NH4Cl
Answer: Option A
: A


On dissolution of NH4Cl into NH3+HCl number of moles become double. Hence, volume is doubled and therefore, density is halved.

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Question 1. 1 mole of N2 O4 (g) at 300K is kept in a closed container under 1 atm pressure. It is heated to 600K when 20% by mass of N2O4 (g) decomposes to NO2 (g) . The resultant pressure is
  1.    1.2 atm
  2.    2.4 atm
  3.    2.0 atm
  4.    1.0 atm
Answer: Option B
: B

Correct option is b).
N2O42NO2Initialmoles10

Total moles=0.8+0.4=1.2

Applying, PV=nRT, at constant volume,

P1n1T1=Pn2T2

11X300=P21.2X600

P2=2.4atm.
Question 2. At 30C,Kp for the dissociation reaction:

SO2Cl2 (g) SO2 (g) + Cl2 (g) is 2.9x 102~atm. If the total pressure is 1 atm, the degree of dissociation of SO2Cl2 is (assume 1α2 =1)
  1.    85%
  2.    12%
  3.    17%
  4.    35%
Answer: Option C
: C

Correct option is c).
SO2Cl2(g)SO2(g)+Cl2(g)1001ααα
Total moles = 1+α
Pso2Cl2=1α/1+α,Pso2=α/1+α,Pcl2=α/1+α
Kp=(α/1+α)2/(1α/1+α)=α2/1α2α2
α=Kp=(2.9×102=0.17,i.e.,17%
Question 3. If concentration of OHions in the reaction  Fe(OH)3(s)  Fe3++(aq)+ 3 OH (aq) is decreased by 1/4 times, then equilibrium concentration of Fe3+will increase by
  1.    8 times
  2.    16 times
  3.    64 times
  4.    4 times.
Answer: Option C
: C


K=[Fe3+][OH]3 If concentration of OHion is decreased to 14 th to keep K constant
Concentration of Fe3+ions has to be increased by 64 times.
Question 4. In a reaction A+2B  2C; 2 moles ofA, 3 mole of B and 2 mole of C are placed in a 2 L flask and the equilibrium concentration of C is 0.5 mole/L. The equilibrium constant K for the reaction is:
  1.    0.073
  2.    0.147
  3.    0.05
  4.    0.026
Answer: Option C
: C


A+2B2CInitialmoles232Molarconc.2/2=1mol/L3/2=1.5mol/L2/2=1mol/LAtEqui.Conc1+0.25=1.25mol/L1.5+0.5=2mol/L0.5mol/L

K=(0.5)2/(1.25×(2)2)=0.05 Note that concentration of C has decreased, i.e., to attain equilibrium, reaction has shifted backwards. Decrease in conc. of C will be 0.5mol/L and hence increase in concentration of B will be 0.5mol/L and that of A will be 0.25mol/L
Question 5. At a certain temperature and a total pressure of 105 Pa, iodine vapours contain 40 % by volume of iodine atoms

[I2(g)  2 I(g)]. Kp for the equilibrium will be
  1.    0.67
  2.    1.5
  3.    2.67 × 104
  4.    9.0 × 104
Answer: Option C
: C

Partial pressure of I atoms
(pl)=40×105/100Pa=0.4×105Pa

Partial pressure of l2(p2)=60×105/100Pa=0.6×105Pa
Kp=(p1)2/p2=(0.4×105)2/(0.6×105)=2.67×104

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