Preparing for the National Eligibility-cum-Entrance Test (NEET) is a challenging journey that demands dedication and practice. Two crucial components of the NEET chemistry section are inorganic and physical chemistry. Mock tests are a powerful tool to help you succeed in these areas.
Inorganic Chemistry: Inorganic chemistry often scares students with its vast array of elements, compounds, and reactions. Mock tests provide a structured approach to this subject. They enable you to gauge your understanding of concepts such as the periodic table, chemical bonding, and coordination compounds. These tests help you identify your weak points, allowing you to focus on them during your revision.
Physical Chemistry: Physical chemistry, on the other hand, delves into the fundamental principles that govern chemical reactions. Mock tests for this section can improve your problem-solving skills, especially in areas like thermodynamics, chemical kinetics, and equilibrium. Regular practice with these tests will boost your confidence and speed in solving numerical problems.
Here are some tips for using mock tests effectively:
Consistency is Key: Take regular mock tests to track your progress.
Analyze Your Mistakes: Carefully review and learn from the questions you answered incorrectly.
Time Management: Practice under timed conditions to improve your speed.
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This test consistes of previous year questions(PYQs) of NEET.
Total duration of this test is 10 minutes.
There is 1 section in the question paper consisting of 10 Single Option Correct MCQ questions.
Each question is allotted 4 (Four) marks for correct response.
No deduction for indicating incorrect response for each question.
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Question 1 of 19
1. Question
4 points
Acetone on treatment with CH3-Mg-I and on further hydrolysis gives
Correct
2-Methyl 2-propanol
Incorrect
2-Methyl 2-propanol
Question 2 of 19
2. Question
4 points
Category: inorganic chemistry
Two oxides of metal contain 27.6% and 30% oxygen respectively. If formula of first oxide is M3O4, then formula of second oxide is:
Correct
Given that, formula of first oxide = M3O4
Let mass of the metal = x % of metal in M3O4 = (3x/ 3x+64) ×100
but as given % age = (100-27.6) = 72.4 % so, (3x/ 3x+64)×100 = 72.4
or x = 56. in 2nd oxide, oxygen = 30%….so metal = 70% so, the ratio is M : O 70/56 : 30/16 1.25 : 1.875 2 : 3 so, 2nd oxide is M2O3
Incorrect
Given that, formula of first oxide = M3O4
Let mass of the metal = x % of metal in M3O4 = (3x/ 3x+64) ×100
but as given % age = (100-27.6) = 72.4 % so, (3x/ 3x+64)×100 = 72.4
or x = 56. in 2nd oxide, oxygen = 30%….so metal = 70% so, the ratio is M : O 70/56 : 30/16 1.25 : 1.875 2 : 3 so, 2nd oxide is M2O3
Question 3 of 19
3. Question
4 points
Category: Physical Chemistry
5.6 litres of a gas at N.T.P. are found to have a mass of 11 g. The molecular mass of the gas is
Correct
Answer:
5.6 liter=11g 22.4 liter=? (x) g x= 22.4×11/5.6 x=44g Thus, Molecular mass of gas is 44g
Incorrect
Answer:
5.6 liter=11g 22.4 liter=? (x) g x= 22.4×11/5.6 x=44g Thus, Molecular mass of gas is 44g
Question 4 of 19
4. Question
4 points
Category: Physical Chemistry
The crystalline salt, Na2SO4.xH2O on heating loses 55.9% of its weight. The formula of the crystalline salt is
Correct
Answer:
Molar mass of salt = (2 × 23) + 32 + (16 × 4) + x × 18 = 142 + 18x
On heating it will lose water and become anhydrous.
55.9% mass is the mass of waater in Na2SO4.xH2O.
Mass by Mass% = (mass / molar mass of compound) × 100 mass of water = 18x 55.9 = (18x / 142 + 18x) × 100 55.9(142 + 18x) = 1800x 7937.8 + 1006.2x = 1800x x = 9.99 x = 10 (approx.) So, the molecular formula of compound is Na2SO4.10H2O.
Incorrect
Answer:
Molar mass of salt = (2 × 23) + 32 + (16 × 4) + x × 18 = 142 + 18x
On heating it will lose water and become anhydrous.
55.9% mass is the mass of waater in Na2SO4.xH2O.
Mass by Mass% = (mass / molar mass of compound) × 100 mass of water = 18x 55.9 = (18x / 142 + 18x) × 100 55.9(142 + 18x) = 1800x 7937.8 + 1006.2x = 1800x x = 9.99 x = 10 (approx.) So, the molecular formula of compound is Na2SO4.10H2O.
Question 5 of 19
5. Question
4 points
Category: inorganic chemistry
0.5 g of a metal on oxidation gave 0.79 g of its oxide. The equivalent weight of the metal is (approximately)
Correct
Given,
Mass of metal = 0.5 g
Mass of oxide = 0.79 g
First we have to calculate the Mass of Oxygen in metal oxide.
Mass of Oxygen = Mass of oxide – Mass of Metal = 0.79 – 0.5 = 0.29 g
The equivalent weight of oxygen is 8 g/eq.
0.29 g of oxygen combines with the 0.5 g of metal
8 g of oxygen combines with of metal.
Therefore, the equivalent weight of metal is 13.793 g/eq.
Incorrect
Given,
Mass of metal = 0.5 g
Mass of oxide = 0.79 g
First we have to calculate the Mass of Oxygen in metal oxide.
Mass of Oxygen = Mass of oxide – Mass of Metal = 0.79 – 0.5 = 0.29 g
The equivalent weight of oxygen is 8 g/eq.
0.29 g of oxygen combines with the 0.5 g of metal
8 g of oxygen combines with of metal.
Therefore, the equivalent weight of metal is 13.793 g/eq.
Question 6 of 19
6. Question
4 points
Category: inorganic chemistry
Air contains nearly 20% oxygen by volume. The volume of air needed for complete combustion of 100 cc of acetylene will be
Correct
2 C2H2 + 5O2 → 4CO2 + 2 H2O
In this reaction, we can see that, 2 moles of C2H2reacts with 5 moles of O2 Or, 1 mole of C2H2 reacts with 5/2 moles of O2 Under standard conditions, we can say that: 22.44 L of C2H2 reacts with (5/2) x 22.4 L moles of O2 So, 100 cc of C2H2 will reacts with (5/2) x 100 cc = 250 cc of O2 .
But, since air contains 20 % of oxygen by volume, the amount of air needed to react with 100 cc of C2H2 will be = 250 x (100/20) = 1250 cc of air.
Incorrect
2 C2H2 + 5O2 → 4CO2 + 2 H2O
In this reaction, we can see that, 2 moles of C2H2reacts with 5 moles of O2 Or, 1 mole of C2H2 reacts with 5/2 moles of O2 Under standard conditions, we can say that: 22.44 L of C2H2 reacts with (5/2) x 22.4 L moles of O2 So, 100 cc of C2H2 will reacts with (5/2) x 100 cc = 250 cc of O2 .
But, since air contains 20 % of oxygen by volume, the amount of air needed to react with 100 cc of C2H2 will be = 250 x (100/20) = 1250 cc of air.
Question 7 of 19
7. Question
4 points
Category: inorganic chemistry
A solution of 10 ml(M/10) FeSO4 was titrated with KMnO4 solution in acidic medium. The amount of KMnO4 used will be
Correct
The equation of the Mn2+ in acidic medium will be-
MnO4– + 8H++ 5 Fe2+ → Mn2++ 5Fe3+ +4H2O.
We can balance the reaction by using redox balancing of acid and base.
So, we know from the equation that 5 moles of iron react to form 1 mole of Mn+2.
Thus, we can say that (10×0.1) moles of iron will react with the (10×0.1/5) moles of Manganese ions, which will be 0.2 thus, 10ml volume of KMnO4 and 0.02M moles of KMnO4. We will get the same result if we use M1V1 = M2V2 formulae.
Incorrect
The equation of the Mn2+ in acidic medium will be-
MnO4– + 8H++ 5 Fe2+ → Mn2++ 5Fe3+ +4H2O.
We can balance the reaction by using redox balancing of acid and base.
So, we know from the equation that 5 moles of iron react to form 1 mole of Mn+2.
Thus, we can say that (10×0.1) moles of iron will react with the (10×0.1/5) moles of Manganese ions, which will be 0.2 thus, 10ml volume of KMnO4 and 0.02M moles of KMnO4. We will get the same result if we use M1V1 = M2V2 formulae.
Question 8 of 19
8. Question
4 points
Category: Physical Chemistry
At S.T.P. the density of CCl4 vapour in g/L will be nearest to
Correct
Molecular Weight of CCl₄ = 35.5 + 12 x 4 = 154 1 mole CCl₄ vapour=154 g According to Avogadro’s hypothesis we know that the molar volume of a gas at STP is 22.4 litres = 154/22.4 = 6.87
Incorrect
Molecular Weight of CCl₄ = 35.5 + 12 x 4 = 154 1 mole CCl₄ vapour=154 g According to Avogadro’s hypothesis we know that the molar volume of a gas at STP is 22.4 litres = 154/22.4 = 6.87
Question 9 of 19
9. Question
4 points
Category: inorganic chemistry
The law of multiple proportions was proposed by
Correct
Dalton’s law of multiple proportions is part of the basis for modern atomic theory, along with Joseph Proust’s law of definite composition (which states that compounds are formed by defined mass ratios of reacting elements) and the law of conservation of mass that was proposed by Antoine Lavoisier.
Incorrect
Dalton’s law of multiple proportions is part of the basis for modern atomic theory, along with Joseph Proust’s law of definite composition (which states that compounds are formed by defined mass ratios of reacting elements) and the law of conservation of mass that was proposed by Antoine Lavoisier.
Question 10 of 19
10. Question
4 points
Category: Physical Chemistry
The anode rays were discovered by
Correct
The discovery of protons can be attributed to Rutherford. In 1886 Goldstein discovered existence of positively charged rays in the discharge tube by using perforated cathode. These rays were named as anode rays or cannal rays.
Incorrect
The discovery of protons can be attributed to Rutherford. In 1886 Goldstein discovered existence of positively charged rays in the discharge tube by using perforated cathode. These rays were named as anode rays or cannal rays.
Question 11 of 19
11. Question
4 points
Category: inorganic chemistry
Which of the following transitions of electrons in the hydrogen, atom will emit maximum energy ?
Correct
energy is given by following formulae,
E = RZ2hc(
1
−
1
)
n12
n22
E will be highest when the transition from n3→n2
Incorrect
energy is given by following formulae,
E = RZ2hc(
1
−
1
)
n12
n22
E will be highest when the transition from n3→n2
Question 12 of 19
12. Question
4 points
Category: inorganic chemistry
The uncertainity found from the uncertainity principle Δx.Δp=h/4 π is
Correct
Δx.Δp ≧ h/4π So by the equation given in the question, the uncertainity value will be the least.
Incorrect
Δx.Δp ≧ h/4π So by the equation given in the question, the uncertainity value will be the least.
Question 13 of 19
13. Question
4 points
Category: inorganic chemistry
The number of orbitals in any shell (n) is equal to
Correct
Maximum number of electrons can be accommodated in a shell of principal quantum number n is 2n2. Since each orbital can accommodate only 2 electrons, maximum number of orbitals =2n2/2= n2
Incorrect
Maximum number of electrons can be accommodated in a shell of principal quantum number n is 2n2. Since each orbital can accommodate only 2 electrons, maximum number of orbitals =2n2/2= n2
Question 14 of 19
14. Question
4 points
Category: inorganic chemistry
The maximum number of 4f electrons having spin quantum number s= (-1/2) is
Correct
Total no. of electrons in F- subshell = 14 (when we include both spin +1/2 & – 1/2) But maximum no. of electrons = 7 if only one spin either – 1/2 or +1/2.
Incorrect
Total no. of electrons in F- subshell = 14 (when we include both spin +1/2 & – 1/2) But maximum no. of electrons = 7 if only one spin either – 1/2 or +1/2.
Question 15 of 19
15. Question
4 points
Category: inorganic chemistry
The number of unpaired electrons in Cu+ (Z= 29) is
Correct
The electronic configuration of Cu is,
1s22s22p63s23p64s13d10
The reason for abnormal electronic configuration of Cu is because of the half-filled and fully-filled configuration are the more stable configuration. In this configuration, the d-subshell is fully-filled and s-subshell is half-filled which makes the Cu more stable.
Now, for getting a positive charge, it has to lose 1 electron which will be removed from the outermost 4s orbital and hence all the remaining electrons will be paired leading to 0 unpaired electrons
Incorrect
The electronic configuration of Cu is,
1s22s22p63s23p64s13d10
The reason for abnormal electronic configuration of Cu is because of the half-filled and fully-filled configuration are the more stable configuration. In this configuration, the d-subshell is fully-filled and s-subshell is half-filled which makes the Cu more stable.
Now, for getting a positive charge, it has to lose 1 electron which will be removed from the outermost 4s orbital and hence all the remaining electrons will be paired leading to 0 unpaired electrons
Question 16 of 19
16. Question
4 points
Category: inorganic chemistry
The electrons present in K-shell of the atom will differ in
Correct
Two electrons in K-shell will differ in spin quantum number. They will have same values for the principal quantum number, azimuthal quantum number, and magnetic quantum number.
For first electron, n=1,l=0,m=0,s=+1/2 For second electron, n=1,l=0,m=0,s=−1/2
Incorrect
Two electrons in K-shell will differ in spin quantum number. They will have same values for the principal quantum number, azimuthal quantum number, and magnetic quantum number.
For first electron, n=1,l=0,m=0,s=+1/2 For second electron, n=1,l=0,m=0,s=−1/2
Question 17 of 19
17. Question
4 points
Category: inorganic chemistry
Consider the following statements. The d-orbitals have
I. four lobes and two nodes
II. four lobes and one node
III. same sign in the opposite lobes
IV. opposite sign in the opposite lobes
Which of the above statements are correct ?
Correct
Total no of angular nodes = l (azimuthal quantum no.), so l=2 in d orbital, therefore number of nodes is 2 and it is a fact that opposite lobes have same signs and adjacent ones have opposite signs. there are 4 lobes in d orbital.
Incorrect
Total no of angular nodes = l (azimuthal quantum no.), so l=2 in d orbital, therefore number of nodes is 2 and it is a fact that opposite lobes have same signs and adjacent ones have opposite signs. there are 4 lobes in d orbital.
Question 18 of 19
18. Question
2 points
Category: Bio-Inorganic Chemistry
The biological functions of the cytochrome P450 and myoglobin are respectively,
Correct
Incorrect
Question 19 of 19
19. Question
2 points
Category: Bio-Inorganic Chemistry
Deoxy-Hemocyanin is,
Correct
<div>
<p><strong>Hemocyanin</strong>, a copper-containing protein chemically unlike hemoglobin, is found in some crustaceans.</p>
<p><strong>Hemocyanin</strong> is B<strong>lue</strong> in <strong>colour</strong> when oxygenated and colourless when oxygen is removed.</p>
</div>
<div></div>
Incorrect
<div>
<p><strong>Hemocyanin</strong>, a copper-containing protein chemically unlike hemoglobin, is found in some crustaceans.</p>
<p><strong>Hemocyanin</strong> is B<strong>lue</strong> in <strong>colour</strong> when oxygenated and colourless when oxygen is removed.</p>
</div>
<div></div>
In summary, NEET mock tests in inorganic and physical chemistry can be your secret weapon to ace the exam. They provide a structured approach to tackle these challenging topics and fine-tune your problem-solving skills. So, incorporate them into your study routine and watch your confidence and performance soar. Good luck with your NEET journey!
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