The West Bengal Joint Entrance Examination Board conducted WBJEE 2018 on April 22 in offline mode. Both papers i.e., Paper I (Mathematics) and Paper II (Physics and Chemistry) were conducted on the same day. A total of 80 questions asked in WBJEE 2018 Paper II. The chemistry section consisted of 40 questions. The Chemistry Question paper was of moderate difficulty level and students with descent preparation could easily attempt 30 to 32 questions. Here, we are providing you the detailed solution of WBJEE Chemistry Question Paper 2018. Students, who are going to appear for WBJEE 2019, must solve this question paper by setting real exam like conditions. After solving the question paper, students can match their answers with the PDF of the solutions provided at the end of the article.
WBJEE 2019: Latest Chemistry Syllabus
Some questions from WBJEE Chemistry Question Paper 2018 are given below:
Question:
The heat of neutralization of a strong base and a strong acid is 13.7 kcal. The heat released when 0.6 mole HCl solution is added to 0.25 mole of NaOH is
(A) 3.425 kcal
(B) 8.22 kcal
(C) 11.645 kcal
(D) 13.7 kcal
Solution:
The following reaction will take place
HCl + NaOH → NaCl + H2O [∆H = –13.7 Kcal (Given)]
(1 Mole) (1 Mole)
HCl + NaOH → NaCl + H2O
(0.25) (0.6)
HCL is limiting reagent then,
Heat released = 13.7 × 0.25 = 3.425 Kcal
Hence, the correct option is (A).
Question:
Which one of the following is a condensation polymer?
(A) PVC
(B) Teflon
(C) Dacron
(D) Polystyrene
Solution:
The condensation polymers are formed by repeated condensation reaction between two different bi-functional or tri-functional monomeric units. The examples are terylene (dacron), nylon 6, 6, nylon 6 etc.
Hence, the correct option is (C).
Question:
Which of the set of oxides are arranged in the proper order of basic, amphoteric, acidic?
(A) SO2, P2O5, CO
(B) BaO, Al2O3, SO2
(C) CaO, SiO2, Al2O3
(D) CO2, Al2O3, CO
Solution:
The basic oxide is BaO.
The amphoteric oxide is Al2O3.
The acidic oxide is SO2
Hence, the correct option is (B).
WBJEE 2019: How to crack the examination in first attempt?
Question:
Out of the following outer electronic configurations of atoms, the highest oxidation state is achieved by which one?
(A) (n – 1) d8 ns2
(B) (n – 1) d5 ns2
(C) (n – 1) d3 ns2
(D) (n – 1) d5 ns1
Solution:
The max number of oxidation states = no. unpaired d-orbital electrons + the 2s electrons.
Therefore, (n – 1) d5 ns2 highest oxidation state of +7.
Hence, the correct option is (B).
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