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JEE Exam  >  JEE Tests  >  Daily Test for JEE Preparation  >  Test: Some Basic Concepts of Chemistry (April 9 ) - JEE MCQ

Test: Some Basic Concepts of Chemistry (April 9 ) - JEE MCQ


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15 Questions MCQ Test Daily Test for JEE Preparation - Test: Some Basic Concepts of Chemistry (April 9 )

Test: Some Basic Concepts of Chemistry (April 9 ) for JEE 2024 is part of Daily Test for JEE Preparation preparation. The Test: Some Basic Concepts of Chemistry (April 9 ) questions and answers have been prepared according to the JEE exam syllabus.The Test: Some Basic Concepts of Chemistry (April 9 ) MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Some Basic Concepts of Chemistry (April 9 ) below.
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Test: Some Basic Concepts of Chemistry (April 9 ) - Question 1
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The number of N atoms in 681 g of C7H5N3O6 is x × 1021. The value of x is ______. (NA = 6.02 × 1023 mol-1) (Nearest Integer)


Detailed Solution for Test: Some Basic Concepts of Chemistry (April 9 ) - Question 1

M.M. of C7H5N3O6 = 84 + 5 + 42 + 96 = 227

Number of N atoms = 9 × 6.02 × 1023
= 5,418 × 1021
∴ The answer is 5,418.

Test: Some Basic Concepts of Chemistry (April 9 ) - Question 2
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Hemoglobin contains 0.34% of iron by mass. The number of Fe atoms in 3.3 g of hemoglobin is: (Given: Atomic mass of Fe is 56 u, NA is 6.022 × 1023 mol-1)

Detailed Solution for Test: Some Basic Concepts of Chemistry (April 9 ) - Question 2

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Test: Some Basic Concepts of Chemistry (April 9 ) - Question 3
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A 2.0 g sample containing MnO2 is treated with HCl liberating Cl2. The Cl2 gas is passed into a solution of KI and 60.0 mL of 0.1 M Na2S2O3 is required to titrate the liberated iodine. The percentage of MnO2 in the sample is ____________ . (Nearest integer)
[Atomic masses (in u) Mn = 55; Cl = 35.5; O = 16, I = 127, Na = 23, K = 39, S = 32]


Detailed Solution for Test: Some Basic Concepts of Chemistry (April 9 ) - Question 3

Test: Some Basic Concepts of Chemistry (April 9 ) - Question 4
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The number of chlorine atoms in 20 mL of chlorine gas at STP is _____ 1021.
(Round off to the nearest integer)
[Assume chlorine is an ideal gas at STP, R = 0.083 L bar mol-1 K-1, NA = 6.023 × 1023]

Detailed Solution for Test: Some Basic Concepts of Chemistry (April 9 ) - Question 4


≈ 1 × 1021

Test: Some Basic Concepts of Chemistry (April 9 ) - Question 5
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On complete combustion of 0.492 g of an organic compound containing C, H and O, 0.7938 g of CO2 and 0.4428 g of H2O was produced. The % composition of oxygen in the compound is ________.(In integer)

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Test: Some Basic Concepts of Chemistry (April 9 ) - Question 6
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Amongst the following statements, that which was not proposed by Dalton was:
Detailed Solution for Test: Some Basic Concepts of Chemistry (April 9 ) - Question 6

Gay Lussac gave the law of gaseous volumes, which states that in the reactions involving gaseous reactants and products, under same conditions of temperature and pressure, the volumes of gases are in the ratio of simple whole numbers.
Therefore, the statement in option (3) is not a postulate of Dalton's atomic theory.

Test: Some Basic Concepts of Chemistry (April 9 ) - Question 7
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Chlorophyll extracted from the crushed green leaves was dissolved in water to make 2 L solution of Mg of concentration 48 ppm. The number of atoms of Mg in this solution is x × 1020 atoms. The value of x is _________. (Nearest Integer)
(Given: Atomic mass of Mg is 24 g mol-1, NA = 6.02 × 1023 mol-1)

Detailed Solution for Test: Some Basic Concepts of Chemistry (April 9 ) - Question 7

Test: Some Basic Concepts of Chemistry (April 9 ) - Question 8
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Two elements A and B form 0.15 moles of A2B and AB3 type compounds. If both A2B and AB3 weigh equally, then the atomic weight of A is ________ times of atomic weight of B. (In integer)

Detailed Solution for Test: Some Basic Concepts of Chemistry (April 9 ) - Question 8

Given: Molar mass of A2B = AB3
∴ (2A + B) = (A + 3B) 
⇒ A = 2B
∴ The atomic weight of A is 2 times the atomic weight of B.
The integer answer is 2.

Test: Some Basic Concepts of Chemistry (April 9 ) - Question 9
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The mole fraction of a solute in a 100 molal aqueous solution _______ × 10-2.
(Round off to the nearest integer)
[Given: Atomic masses: H: 1.0 u, O: 16.0 u]

Detailed Solution for Test: Some Basic Concepts of Chemistry (April 9 ) - Question 9

100 molal aqueous solution means there are 100 moles of solute in 1 kg = 1000 gm of water.
Now, mole fraction of solute = 
=  = 0.6428
= 64.28 × 10–2
≈ 64 × 10-2

Test: Some Basic Concepts of Chemistry (April 9 ) - Question 10
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The complete combustion of 0.492 g of an organic compound containing 'C', 'H' and 'O' gives 0.793 g of CO2 and 0.442 g of H2O. The percentage of oxygen composition in the organic compound is __________. (Nearest integer)

Detailed Solution for Test: Some Basic Concepts of Chemistry (April 9 ) - Question 10

Test: Some Basic Concepts of Chemistry (April 9 ) - Question 11
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Among the following, the aromatic compounds are:
(A) 
(B) 
(C) 
(D) 
Choose the correct answer from the following options:
Detailed Solution for Test: Some Basic Concepts of Chemistry (April 9 ) - Question 11

According to Huckel rule for aromaticity, the molecule must be a planar, cyclic system having delocalised (4n + 2)π electrons, where n is an integer, i.e. 0, 1, 2, or 3. Thus, aromatic compounds have 2, 6, 10 or 14π electrons.

(A) Non-aromatic
(B) Aromatic
(C) Aromatic
(D) Anti-aromatic

Test: Some Basic Concepts of Chemistry (April 9 ) - Question 12
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Two solutions, A and B, each of 100 L was made by dissolving 4 g of NaOH and 9.8 g of H2SO4 in water, respectively. The pH of the resultant solution obtained from mixing 40 L of solution A and 10 L of solution B is ________.
[Answer upto two decimal places]

Detailed Solution for Test: Some Basic Concepts of Chemistry (April 9 ) - Question 12

For given solutions, we have:
Moles of NaOH = 4/40 = 0.1
Moles of H2SO4 = 9.8/98 = 0.1
Molarity of each sample = 0.1/100 = 0.001 M
Now, 40 L of NaOH solution and 10 L of H2SO4 solution are added. Thus, we get:
Total volume = 50 L
Equivalents of NaOH = 0.001 × 40 = 0.04
Equivalents of H2SO4 = 0.001 × 10 × 2 = 0.02
Thus, Eq. of NaOH left = 0.04 - 0.02 = 0.02
[OH-] = 0.02/50 = 4 × 10-4
pOH = -log[4 × 10-4]
pOH = -log4 - log10-4
pOH = -0.602 + 4 = 3.398 ≈ 3.4
Further, we know:
pH = 14 - 3.4
pH = 10.6

Test: Some Basic Concepts of Chemistry (April 9 ) - Question 13
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Complete combustion of 750 g of an organic compound provides 420 g of CO2 and 210 g of H2O. The percentage composition of carbon and hydrogen in organic compound is 15.3 and ________, respectively. (Round off to the nearest integer)

Detailed Solution for Test: Some Basic Concepts of Chemistry (April 9 ) - Question 13

44 gm of CO2 has 12 gm of carbon.
So, 420 gm of CO2 ⇒ (22/44) × 420
⇒ (1260/11) gm of carbon
⇒ 114.545 gm of carbon
So, %age of carbon = (114.545/750) × 100
≌ 15.3%
18 gm of H2O ⇒ 2 gm of H2
210 gm ⇒ (2/18)  × 210
= 23.33 gm of H2
So, %age of H2 ⇒ (23.33/750) × 100 = 3.11%
≈ 3%

Test: Some Basic Concepts of Chemistry (April 9 ) - Question 14
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Consider the above reaction, the limiting reagent of the reaction and the number of moles of NH3 formed respectively are:
Detailed Solution for Test: Some Basic Concepts of Chemistry (April 9 ) - Question 14

Test: Some Basic Concepts of Chemistry (April 9 ) - Question 15
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56 L of nitrogen gas is mixed with excess of hydrogen gas and it is found that 20 L of ammonia gas is produced. The volume of unused nitrogen gas is found to be ____ L.

Detailed Solution for Test: Some Basic Concepts of Chemistry (April 9 ) - Question 15

N2(g) + 3H2(g) → 2NH3(g)
Since H2 is in excess and 20L of ammonia gas is produced.
Hence, 2 Moles NH3 ≡ 1mole N2 (v ∝ n)
20L NH3 ≡ 10 LN2
Volume of N2 left = 56 - 10
= 46 L

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