Punjab Board Class 12 Chemistry Sample Paper with Answer Exam 2025
1. An unripe mango placed in a concentrated salt solution to prepare pickle, shrivels because-
A. It gains water due to osmosis.B. It loses water due to osmosis.
C. It gains water due to reverse osmosis.
D. It loses water due to reverse osmosis.
Answer
It loses water due to osmosis.
2. In comparison to a 0.01 M solution of glucose, the depression in freezing point of a 0.01 M MgCl2 solution is?
A. The sameB. About six
C. About three times
D. About twice times
Answer
About three times.
3. An Isotonic solution have
A. same boiling pointB. Same osmotic Pressure
C. same vapour Pressure
D. Same melting point
Answer
Same osmotic Pressure.
4. What is the final oxidation state of manganese after the electrochemical reactions in a Dry Cell?
A. +1B. +2
C. +3
D. +4
Answer
+3.
5. If the unit of specific rate constant (k) for a certain gaseous reaction is atom-2s-1, then, the order of the reaction is-
A. Zero orderB. First order
C. Second order
D. Third order
Answer
Third order.
6. The coordination number of platinum in [PtCl2(C5H5N)(NH3)] is-
A. 3B. 4
C. 5
D. 6
Answer
4.
7. The reaction of toluene with Cl2 in the presence of FeCl3 gives predominantly-
A. benzoyl chlorideB. benzyl chloride
C. m-chlorotoluene
D. o- and p- chlorotoluene
Answer
o- and p- chlorotoluene.
8. Which of the following is most reactive towards nucleophilic addition reactions?
A. CH3COCH3B. CH3CHO
C. CH3COCH2CH3
D. HCHO
Answer
HCHO.
9. Which of the following reagents cannot be used to distinguish between pentanal and 2-pentanone?
A. Tollen's reagentB. Fehling's Solution
C. Br2 in CCl2
D. I2 in NaOH
Answer
Br2 in CCl2.
10. Which of these is most acidic?
A. CF3COOHB. CCl3COOH
C. CBr3COOH
D. CH3COOH
Answer
CF3COOH.
True/False
11. The compounds [CoCl2(NH3)4] NO2 and [CoCl(NO2)(NH3)4)Cl show coordination isomerism.Answer
TRUE.
12. The crystal field splitting !::J.0, depends on the field produced by the ligand and charge on the metal ion.
Answer
TRUE.
13. The boiling point of ethers are higher than those of isomeric alcohols.
Answer
False.
14. Benzaldehyde cannot undergo Cannizzaro reaction.
Answer
TRUE.
15. The red brown precipitate of Aldehydes with Fehling's solution is due to the formation of Cu2O
Answer
TRUE.
Read the passage and answer the questions 16 to 20-
Carbohydrates are optically active polyhydroxy a/dehydes and ketones or those compounds which on hydrolysis give such compounds are also carbohydrates. The carbohydrates which are not hydrolysed are called monosaccharides.Monosaccharides with a/dehydic group are called A/doses and those with free Ketonic group are called Ketoses. Carbohydrates are optically active. Number of optical isomers= 2 n, where n= number of asymmetric carbons. Carbohydrates are mainly synthesised by plants during photosynthesis. The monosaccharides exist in the form of cyclic structures. In cyclization, the -OH group combines with the aldehydic or ketonic group. As a result, cyclic structures of five or six membered rings containing one oxygen are formed e.g. Glucose, Fructose, Galactose.
16. What are carbohydrates?17. What are Aldoses?
18. Define Monosaccharides.
19. Name a monosaccharide.
20. Glucose molecule has four asymmetric carbons. Find the total number of optical isomers in glucose.
16. What are carbohydrates?
Answer
Carbohydrates are optically active polyhydroxy aldehydes and ketones, or compounds that yield such compounds upon hydrolysis.
Example: monosaccharides which cannot be further hydrolyzed.
17. What are Aldoses?
Answer
Aldoses are monosaccharides that contain an aldehydic group (-CHO). They are characterized by having a carbonyl group at the end of the carbon chain.
18. Define Monosaccharides.
Answer
Monosaccharides are the simplest carbohydrates that cannot be hydrolyzed into smaller sugars. They are either polyhydroxy aldehydes or ketones, generally containing two or more hydroxyl (-OH) groups.
19. Name a monosaccharide.
Answer
Glucose is an example of a monosaccharide.
20. Glucose molecule has four asymmetric carbons. Find the total number of optical isomers in glucose.
Answer
The total number of optical isomers can be calculated by using the formula 2n2, where n is the number of asymmetric (chiral) carbons.
Glucose has four asymmetric carbons, so the-
Total optical isomers = 24 = 16
Thus, glucose can have 16 optical isomers.
