CBSE 12TH CHEMISTRY - Online Test
Q1. Enzymes are
Answer : Option A
Explaination / Solution:
Enzymes are biological catalyst, they catalize different chemical reactions in body.
Q2. Honey belongs to one of the following category
Answer : Option D
Explaination / Solution:
Honey is a carbohydrate
Q3. benzaldehye differs from acetaldehyde in that
Answer : Option C
Explaination / Solution:
Fehling's Solution Test: When acetaldehyde is heated with Fehling solution, red ppt. of cuprous oxide are formed. But benzaldehyde does not give red ppt. with fehling solution.
Reaction are
CH3CHO + 2Cu2+ + 5OH-
CH3COO- + Cu2O + 3H2O
C6H5CHO + 2Cu2+ + 5OH-
Q4.
Arrange the following in decreasing order of their basic strength :
Answer : Option B
Explaination / Solution:
Alkyl amines and ammonia are more basic than aniline because in aniline lone pair on N is in resonace with benzene ring so is less available.
Also more 20 alkyl amines are more basic than primary amine and ammonia due to +I effect of the alkyl group which increases basicity.
Q5. Solubility of gas decreases with
Answer : Option C
Explaination / Solution:
Because of increase in kinetic energy of gas molecules with temperature there tendency to escape from the liquid will increase.
Q6. Addition polymers formed by the polymerisation of a single monomeric species
Answer : Option C
Explaination / Solution:
When single monomer is present polymer is homopolymer.
Q7.
is the structure of
is the structure of
Answer : Option C
Explaination / Solution:
H3PO2 is Hypophosphorus acid. It has one P-OH, two P-H and one P=O bonds.
Q8. Conversion of ore to oxide is generally carried out by the process of
Answer : Option B
Explaination / Solution:
Calcination is employed for carbonate ores to convert into respective oxide ores. Roasting is employed for sulphide ores for convertion to oxide ores.
Q9. Which one of the following compound is deliquescent?
Answer : Option A
Explaination / Solution:
ZnCl2 is deliquescent.
Q10. Arrange the following sets of compounds in order of their increasing boiling points : Pentan – 1 – ol, butan – 1 – ol, butan – 2 – ol, ethanol, propan – 1 – ol, methanol
Answer : Option B
Explaination / Solution:
Boiling point is α Vanderwall forces α Molecular weight.
