CBSE 11TH CHEMISTRY - Online Test
Q1. For an isolated system, ΔU = 0, what will be ΔS?
Answer : Option A
Explaination / Solution:
for isolated system ΔS = 0.
Q2.
Hybridization of and of are
Answer : Option A
Explaination / Solution:
C2 is sp2 while C3 is sp (special case of allenes)
Q3. Beryllium carbonate is unstable and can be kept only in the atmosphere of
Answer : Option C
Explaination / Solution:
is very unstable due to size incompatibility.
Q4.
Sodium cyanide, sulphide and halide,( -CN, -S and -H, coming from organic compound) so formed on sodium fusion are extracted from the fused mass by boiling it with distilled water. This is called
Answer : Option A
Explaination / Solution:
This is lassaignes test.
Q5. The van der Waals Equation adjusts the measured pressure
Answer : Option B
Explaination / Solution:
/ is added.
Q6.
E0 values of some redox couples are given below. On the basis of these values choose the correct option.
E0 values: Br2/Br- = +1.90; Ag+/Ag = +0.80; Cu2+/Cu = +0.34; I2/I-(s) = +0.54
Answer : Option C
Explaination / Solution:
Because in electrochemical series Cu is above all in sequence I2, Ag and Br2, therefore, Cu will reduce Br2.
Q7. What is the mass of 1 molecule of ?
Answer : Option B
Explaination / Solution:
Since ,as per Avogadro's law ,
the mass of 6.02 x molecules of
= 32 g
mass of 1 molecule of O2
=
= 5.32 x g
Q8. At 500 K, equilibrium constant, , for the following reaction is 5.2. HI (g) What would be the equilibrium constant Kc for the reaction 2HI (g)
Answer : Option B
Explaination / Solution:
The second equation can be obtained by reversing the 1st reaction and multiplying by 2. So Here ,=5.2
Q9. Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of
Answer : Option D
Explaination / Solution:
The presence of double bond on terminal carbon will give formaldehyde as one of the products.
Q10. Choose one of the following in the increasing order of electron gain enthalpy
Answer : Option C
Explaination / Solution:
In general as the we move down the group electron gain enthalpy decreases; so the expected trend should be I < Br < Cl < F but in actual the order is I < Br < F < Cl . This is due to small size of F atom. The valence electrons of F atom provide strong repulsion to incoming electron, hence F has unexpectedly low electron affinity.
