All Courses
Get the App Signup / Login
Sign in Open App
Class 9 Exam  >  Class 9 Tests  >  Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Class 9 MCQ

Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Class 9 MCQ


Test Description

30 Questions MCQ Test - Atoms And Molecules - Olympiad Level MCQ, Class 9 Science

Atoms And Molecules - Olympiad Level MCQ, Class 9 Science for Class 9 2024 is part of Class 9 preparation. The Atoms And Molecules - Olympiad Level MCQ, Class 9 Science questions and answers have been prepared according to the Class 9 exam syllabus.The Atoms And Molecules - Olympiad Level MCQ, Class 9 Science MCQs are made for Class 9 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science below.
Solutions of Atoms And Molecules - Olympiad Level MCQ, Class 9 Science questions in English are available as part of our course for Class 9 & Atoms And Molecules - Olympiad Level MCQ, Class 9 Science solutions in Hindi for Class 9 course. Download more important topics, notes, lectures and mock test series for Class 9 Exam by signing up for free. Attempt Atoms And Molecules - Olympiad Level MCQ, Class 9 Science | 35 questions in 25 minutes | Mock test for Class 9 preparation | Free important questions MCQ to study for Class 9 Exam | Download free PDF with solutions
Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 1
Save

The law of constant proportions was proposed by:

Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 1

The law of constant proportion was given by Joseph Proust in 1779. This observation was first made by the English theologian and chemist Joseph Priestley, and Antoine Lavoisier, a French nobleman and chemist centered on the process of combustion.

Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 2
Save

The weight of two elements which combine with each other are in the ratio of their :-

Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 2
The equivalent weight of any substance is the mass of the substance in grams that combines with or is chemically equivalent to eight grams of oxygen or one gram of hydrogen. In other words, it is equal to the mass of the substance in grams that would react with or replace one gram of hydrogen. Equivalent weight is expressed as the ratio of atomic weight or molecular weight to the valence.The n number of equivalents is the part of this equation is the quantity of charge in an equation, and what charge is used is depends on the scenario.
1 Crore+ students have signed up on EduRev. Have you? Download the App
Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 3
Save

How many moles of oxygen atoms are present in one mole of acetic acid?

Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 3
Formula of ACETIC ACID is CH3COOH. so from the chemical formula, you get that number of oxygen atoms in a molecule of acetic acid=2so, 1 mol of acetic acid will contain 2 moles of oxygen atomshope.
Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 4
Save

What is the number of particles in one mole of a substance?
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 4
1 mole of any substance contains 6.022 x 1023 particles.
6.022 x 1023 is known as the Avogadro Number or Avogadro Constant and is given the symbol NA(1)
N = n x NA
N = number of particles in the substanc
n = amount of substance in moles (mol)
NA = Avogardro Number = 6.022 x 1023 particles mol-1
N = n x (6.022 x 1023)
To find the number of particles, N, in a substance:
N = n x NA
N = n x (6.022 x 1023)
To find the amount of substance in moles, n :
n = N x NA
n = N x (6.022 x 1023)
Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 5
Save

How many atoms and how many gram atoms are there in 10 grams of calcium ?
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 5

No.of particles = mass/ molecular mass x avagadro constant

mass of calcium = 10 g
molecular mass of calcium = 40 u
therefore no of particles = 10/40 x 6.022 x 1023
= 1.5055 x 1023.

=> Gram atoms is mole which is equal to =10/40
=0.25

Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 6
Save

Calculate the weight of 0.1 mole of sodium carbonate :-
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 6
moles = mass in gm/mol.wt of compound

0.1 moles of sodium carbonate=wt. of Na2CO3/1O6

wt. of Na2CO3=10.6 gm
Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 7
Save

How many number of moles are present in 540 g of glucose?
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 7
Molecular weight of glucose =180g 
Thus, 180g =1 mole
So, 540 g= 540/180=3 moles.
Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 8
Save

Formula for Aluminium Oxide is:
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 8
Formula for Aluminium Oxide:
The correct formula for Aluminium Oxide is Al2O3.
Explanation:
Aluminium Oxide is a chemical compound composed of aluminium and oxygen atoms. The formula Al2O3 indicates that there are two aluminium atoms and three oxygen atoms present in one molecule of aluminium oxide.
Here's a breakdown of the formula:
- Al: This represents the symbol for the element aluminium.
- 2: This subscript number indicates the number of aluminium atoms present in one molecule of aluminium oxide.
- O: This represents the symbol for the element oxygen.
- 3: This subscript number indicates the number of oxygen atoms present in one molecule of aluminium oxide.
Therefore, the formula Al2O3 correctly represents the composition of aluminium oxide.
Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 9
Save

Chemical analysis of a carbon compound gave the following percentage composition by weight of the elements
present in it. Carbon = 10.06%, hydrogen = 0.84%, chlorine = 89.10%. Calculate the empirical formula of the compound :-
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 9

To determine the empirical formula of the compound, we need to find the simplest whole number ratio of the elements present in it.
1. Convert the weight percentages of each element into grams:
- Carbon: 10.06% → 10.06g
- Hydrogen: 0.84% → 0.84g
- Chlorine: 89.10% → 89.10g
2. Find the moles of each element by dividing the mass (in grams) by its atomic weight:
- Carbon: 10.06g / 12.01 g/mol = 0.838 mol
- Hydrogen: 0.84g / 1.01 g/mol = 0.832 mol
- Chlorine: 89.10g / 35.45 g/mol = 2.51 mol
3. Divide each mole value by the smallest mole value:
- Carbon: 0.838 mol / 0.832 mol ≈ 1
- Hydrogen: 0.832 mol / 0.832 mol = 1
- Chlorine: 2.51 mol / 0.832 mol ≈ 3
4. The ratios obtained represent the subscripts in the empirical formula. Therefore, the empirical formula of the compound is CHCl3.
Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 10
Save

0.202 g of a carbon compound, on combustion, gave 0.361 g of carbon dioxide and 0.47 g of water. Calculate
the percentage composition of carbon :-
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 10
Given,

Mass of carbon compound = 0.202 g

Mass of carbon dioxide = 0.361 g

Mass of water = 0.47 g

Molar mass of carbon = 12 g/mole

Molar mass of Carbon dioxide = 44 g/mole

First we have to calculate the moles of .

Moles of  = 

The moles of  = 0.008205 moles

In , there are 1 carbon atom and 2 atom molecules are present.

So, the moles of  is equal to the moles of .

The moles of  = 0.008205 moles

Now we have to calculate the mass of Carbon.

Mass of Carbon = Moles of Carbon X Molar mass of Carbon

Mass of Carbon = 0.008205 moles X 12 g/mole = 0.09846 g

Now we have to calculate the % composition of Carbon.


Therefore, the % composition of Carbon is 48.743%.
Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 11
Save

One mole of CO2 contains :
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 11
Explanation:
- One mole of CO2 contains:
- A: 1 gram atom of carbon: This is because the molar mass of carbon is 12 grams/mol, and since there is only 1 carbon atom in CO2, one mole of CO2 will contain 1 gram atom of carbon.
- B: 2 gram atoms of oxygen: This is because the molar mass of oxygen is 16 grams/mol, and since there are 2 oxygen atoms in CO2, one mole of CO2 will contain 2 gram atoms of oxygen.
- C: 3 gram atoms of carbon and oxygen: This is the sum of the gram atoms of carbon and oxygen mentioned above, which is 1 gram atom of carbon + 2 gram atoms of oxygen = 3 gram atoms of carbon and oxygen.
- D: All of the above: This is the correct answer because all the statements A, B, and C are true.
- Therefore, one mole of CO2 contains 1 gram atom of carbon, 2 gram atoms of oxygen, and 3 gram atoms of carbon and oxygen.
Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 12
Save

The percentage of nitrogen in ammonia is given by the expression :-
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 12

To find the percentage of nitrogen in ammonia, we need to calculate the ratio of the atomic mass of nitrogen to the molecular mass of ammonia and then multiply by 100.
The atomic mass of nitrogen is 14 g/mol, and the molecular mass of ammonia is 17 g/mol. Therefore, the percentage of nitrogen in ammonia is given by the expression:
Percentage of Nitrogen = (Atomic mass of nitrogen / Molecular mass of ammonia) x 100
To simplify the expression, we can substitute the values of the atomic mass of nitrogen and the molecular mass of ammonia:
Percentage of Nitrogen = (14 / 17) x 100
Calculating this expression will give us the percentage of nitrogen in ammonia. Let's simplify it further:
Percentage of Nitrogen = (14 / 17) x 100
Percentage of Nitrogen = 0.8235 x 100
Percentage of Nitrogen = 82.35%
Therefore, the correct option is A: 14 x 100/17, which gives us the percentage of nitrogen in ammonia as 82.35%.
Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 13
Save

The empricial formula of a compound is CH2O. Its molecular weight is 90. Calculate the molecular formula of
the compound. (Atomic weight C = 12, H = 1, O = 16)
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 13

To determine the molecular formula of the compound, we need to find the ratio between the empirical formula and the molecular weight.
1. Calculate the empirical formula weight:
- Carbon (C): 12 * 1 = 12
- Hydrogen (H): 1 * 2 = 2
- Oxygen (O): 16 * 1 = 16
Total empirical formula weight: 12 + 2 + 16 = 30
2. Calculate the ratio between the empirical formula weight and the molecular weight:
Molecular weight: 90
Ratio: 90 / 30 = 3
3. Multiply the subscripts in the empirical formula by the ratio:
- Carbon (C): 1 * 3 = 3
- Hydrogen (H): 2 * 3 = 6
- Oxygen (O): 1 * 3 = 3
4. Write the molecular formula using the new subscripts:
C3H6O3
Therefore, the molecular formula of the compound is C3H6O3 (Option C).
Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 14
Save

Calculate the weight of 2.5 mole of CaCO3 :-
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 14

To calculate the weight of 2.5 moles of CaCO3, we need to use the concept of molar mass.
The molar mass of CaCO3 can be calculated by adding the atomic masses of its constituent elements: Ca (calcium), C (carbon), and O (oxygen).
The atomic masses are as follows:
- Ca: 40.08 g/mol
- C: 12.01 g/mol
- O: 16.00 g/mol
Now, let's calculate the molar mass of CaCO3:
Molar mass of CaCO3 = (1 * Ca) + (1 * C) + (3 * O)
= (1 * 40.08 g/mol) + (1 * 12.01 g/mol) + (3 * 16.00 g/mol)
= 40.08 g/mol + 12.01 g/mol + 48.00 g/mol
= 100.09 g/mol
To find the weight of 2.5 moles of CaCO3, we can use the formula:
Weight = number of moles * molar mass
Substituting the values:
Weight = 2.5 moles * 100.09 g/mol
= 250.225 g
Rounding off to the nearest gram, the weight of 2.5 moles of CaCO3 is approximately 250 g.
Therefore, the correct answer is A: 250 g.
Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 15
Save

How many gram atoms are present in 256 g of O2 ?
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 15

To determine the number of gram atoms present in 256 g of O2, we need to follow these steps:
1. Find the molar mass of O2:
- The molar mass of oxygen (O) is 16 g/mol.
- Since there are two oxygen atoms in O2, the molar mass of O2 is 2 * 16 g/mol = 32 g/mol.
2. Determine the number of moles of O2 in 256 g:
- Number of moles = mass / molar mass = 256 g / 32 g/mol = 8 mol.
3. Convert moles to gram atoms:
- 1 mole of any substance contains Avogadro's number of particles, which is approximately 6.022 x 1023 particles.
- Therefore, 8 moles of O2 contain 8 * (6.022 x 1023) gram atoms = 4.8176 x 1024 gram atoms.
4. Write the answer in scientific notation:
- The answer is 4.8176 x 1024 gram atoms.
Therefore, the correct answer is option D: 9.63 x 1024.
Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 16
Save

The molecular formula of potassium nitrate is ________.
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 16
Molecular Formula of Potassium Nitrate

The molecular formula of potassium nitrate is KNO3.


Detailed Solution

To determine the molecular formula of potassium nitrate, we need to consider the elements present in the compound and their respective numbers.


Potassium nitrate consists of the following elements:



  • Potassium (K)

  • Nitrogen (N)

  • Oxygen (O)


We can determine the molecular formula by observing the valency of each element and balancing the charges.


Potassium has a valency of +1, nitrogen has a valency of -3, and oxygen has a valency of -2.


To balance the charges, we need three potassium ions, one nitrogen ion, and three oxygen ions. This gives us the formula KNO3.


Therefore, the correct molecular formula for potassium nitrate is KNO3.

Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 17
Save

The number of moles present in 20 grams of CaCO3 is :-
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 17

No, i think the answer is no. of moles =actual mass/ molecular weight no. of moles=20/41+12+16*3 no. of moles=20/101 no. of moles=0.19801

Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 18
Save

A hydrogen contain 90% of carbon and 10% hydrogen. The empirical formula of the compound is :-
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 18
Given Information:
- The compound contains 90% carbon and 10% hydrogen.
- We need to find the empirical formula of the compound.

To find the empirical formula, we need to determine the ratio of the elements in the compound.
Step 1: Convert the percentages to grams.
- Assume we have 100 grams of the compound.
- Carbon: 90% of 100 grams = 90 grams
- Hydrogen: 10% of 100 grams = 10 grams
Step 2: Convert the grams to moles.
- Carbon: 90 grams / molar mass of carbon (12 g/mol) = 7.5 moles
- Hydrogen: 10 grams / molar mass of hydrogen (1 g/mol) = 10 moles
Step 3: Find the simplest whole number ratio of the elements.
- Divide the number of moles of each element by the smallest number of moles.
- Carbon: 7.5 moles / 7.5 moles = 1
- Hydrogen: 10 moles / 7.5 moles = 1.33
Step 4: Adjust the ratio to get whole numbers.
- Multiply the ratio by the smallest whole number that will make both elements whole numbers.
- Carbon: 1 x 3 = 3
- Hydrogen: 1.33 x 3 = 4
Step 5: Write the empirical formula.
- The empirical formula of the compound is C3H4.
Conclusion:
The empirical formula of the compound is C3H4, so the correct answer is C.
Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 19
Save

An element has only one type of :-
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 19
An element has only one type of:
Atoms:
- Atoms are the basic building blocks of matter.
- Each element consists of atoms that have the same number of protons in their nucleus.
- The number of protons determines the identity of the element.
- Elements are represented by their atomic symbol, such as H for hydrogen or O for oxygen.
- Different elements have different properties due to the unique arrangement of their atoms.
Molecules:
- Molecules are formed when two or more atoms chemically combine.
- While elements consist of only one type of atom, molecules consist of multiple atoms of the same or different elements.
- Molecules can be composed of atoms of the same element, such as O2 (oxygen gas), or atoms of different elements, such as H2O (water).
Mixtures:
- Mixtures are combinations of two or more substances that are physically combined and can be separated by physical means.
- Unlike elements, mixtures do not have a fixed composition or a specific ratio of components.
- Mixtures can be homogeneous (uniformly mixed) or heterogeneous (not uniformly mixed).
Solutes:
- Solutes are substances that are dissolved in a solvent to form a solution.
- While elements can exist in various states (solid, liquid, gas), solutes are typically in a dissolved state.
- Solutes can be elements or compounds.
Conclusion:
- The correct answer is atoms because an element consists of only one type of atom.
Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 20
Save

The simplest formula of a compound having 50% of X(atomic weight 10) and 50% of Y(atomic weight 20) is :-
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 20

To find the simplest formula of the compound, we need to determine the ratio of the elements X and Y in the compound based on their atomic weights.
1. Determine the atomic weight ratio:
- X has an atomic weight of 10.
- Y has an atomic weight of 20.
- The compound is composed of 50% X and 50% Y.
- Therefore, the atomic weight ratio of X to Y is 10:20 or 1:2.
2. Write the simplest formula based on the atomic weight ratio:
- The simplest formula is written by using subscripts to represent the number of atoms of each element in the compound.
- Since the ratio of X to Y is 1:2, we can write the formula as X2Y.
- The subscript "2" indicates that there are two atoms of X in the compound, and the subscript "1" is not explicitly written.
3. Compare the options provided:
- Option A: XY - This formula does not match the atomic weight ratio of 1:2.
- Option B: X2Y - This formula matches the atomic weight ratio of 1:2.
- Option C: XY3 - This formula does not match the atomic weight ratio of 1:2.
- Option D: X2Y2 - This formula suggests that there are two atoms of X and two atoms of Y, which does not match the atomic weight ratio of 1:2.
Therefore, the correct answer is B: X2Y, which represents the simplest formula of the compound with 50% X and 50% Y based on their atomic weights.
Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 21
Save

How many moles of glucose (C6H12O6) are present in 5.4 g ?
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 21

molecular wt of glucose C6H12O6

6*12+12*1+6*16

72+12+96

180

no of moles = wt/mol wt

=5.4/180

=54/1800

=27/900

=9/300

=3/100

=0.03

Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 22
Save

Calculate the number of gram atoms present in 8g of helium
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 22

To calculate the number of gram atoms present in 8g of helium, we need to follow the steps below:
Step 1: Determine the molar mass of helium.
- The molar mass of helium (He) is approximately 4 g/mol.
Step 2: Convert the given mass to moles.
- We can use the formula: Moles = Mass / Molar mass.
- Moles = 8g / 4 g/mol = 2 mol.
Step 3: Calculate the number of gram atoms.
- The number of gram atoms is equal to the number of moles multiplied by Avogadro's number.
- Number of gram atoms = Moles * Avogadro's number.
- Avogadro's number is approximately 6.022 x 10^23.
- Number of gram atoms = 2 mol * 6.022 x 10^23 = 1.2044 x 10^24 gram atoms.
Therefore, there are approximately 1.2044 x 10^24 gram atoms present in 8g of helium.
Final Answer: B. 4
Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 23
Save

How many moles are present in 5.3 g of anhydrous sodium carbonate ?
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 23
Moles of Na2CO3=5.3/106=0.05
so moles of Na=0.05x2=0.1
moles of C=0.05x1=0.05
moles of O=0.05x3=0.15
we can now find the no. of atoms of each type by multiplying the no. of moles by Avagadro's constant.

Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 24
Save

Calculate the number of moles in 60 gram NaOH.
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 24
According to me, no. of moles = given mass/molecular mass no. of moles = 60/23 +16+ 1 no. of moles = 60/40 no. of moles = 3/2 no. of moles = 1.5
Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 25
Save

Calculate the weight of nitrogen present in 0.5 moles of NH3.
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 25

Nitrogen present in 1 mole ammonia = 1 mole
= 14g
So nitrogen present in 0.5 mole ammonia = 14g × 0.5

= 7g

Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 26
Save

Calculate the number of moles present in 7.3 g of HCl.
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 26

To calculate the number of moles present in 7.3 g of HCl, we need to use the formula:
Number of moles = Mass / Molar mass
1. Find the molar mass of HCl:
- H (hydrogen) has a molar mass of approximately 1 g/mol.
- Cl (chlorine) has a molar mass of approximately 35.5 g/mol.
- So, the molar mass of HCl is 1 + 35.5 = 36.5 g/mol.
2. Substitute the values in the formula:
- Number of moles = 7.3 g / 36.5 g/mol
- Number of moles = 0.2 mol
Therefore, there are 0.2 moles present in 7.3 g of HCl.
Answer: D. 0.2
Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 27
Save

Calculate the weight in gram of 0.9 gram atoms of zinc.
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 27
Calculation:
To calculate the weight in grams of 0.9 gram atoms of zinc, we need to use the atomic mass of zinc.
Step 1:
Find the atomic mass of zinc. The atomic mass of zinc is 65.38 g/mol.
Step 2:
Convert the number of gram atoms to moles. To do this, divide the given number of gram atoms (0.9) by Avogadro's number (6.022 x 10^23).
Step 3:
Multiply the number of moles by the atomic mass of zinc to get the weight in grams.
Detailed
Given: 0.9 gram atoms of zinc
Step 1:
Atomic mass of zinc: 65.38 g/mol
Step 2:
Number of moles:
0.9 gram atoms ÷ 6.022 x 10^23 = 0.149 mol
Step 3:
Weight in grams:
0.149 mol x 65.38 g/mol = 9.84 g
Therefore, the weight in grams of 0.9 gram atoms of zinc is approximately 9.84 g.
The correct answer is not listed in the options.
Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 28
Save

Calculate the weight of 0.4 gram atoms of carbon
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 28

To calculate the weight of 0.4 gram atoms of carbon, we need to use the concept of molar mass and Avogadro's number.
Step 1: Find the molar mass of carbon
The molar mass of carbon (C) is 12.01 g/mol. This means that one mole of carbon weighs 12.01 grams.
Step 2: Convert grams to moles
To convert grams to moles, we divide the given mass by the molar mass.
Number of moles = mass (g) / molar mass (g/mol)
For 0.4 grams of carbon:
Number of moles = 0.4 g / 12.01 g/mol = 0.033 moles
Step 3: Convert moles to gram atoms
To convert moles to gram atoms, we multiply the number of moles by Avogadro's number.
Number of gram atoms = number of moles * Avogadro's number
Avogadro's number is approximately 6.022 x 10^23.
For 0.033 moles of carbon:
Number of gram atoms = 0.033 moles * 6.022 x 10^23 = 1.99 x 10^22 gram atoms
Step 4: Calculate the weight
To calculate the weight, we multiply the number of gram atoms by the molar mass.
Weight = number of gram atoms * molar mass
For 1.99 x 10^22 gram atoms of carbon:
Weight = 1.99 x 10^22 * 12.01 g/mol = 2.39 x 10^23 grams
Step 5: Convert grams to grams
To convert grams to grams, we divide the weight by 1,000.
Weight = 2.39 x 10^23 grams / 1,000 = 2.39 x 10^20 grams
Therefore, the weight of 0.4 gram atoms of carbon is approximately 2.39 x 10^20 grams, which is equivalent to 2.39 x 10^17 grams.
Answer: B. 4.8 g
Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 29
Save

What is the weight of 3-gram atoms of sulphur?
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 29

Weight of atom = atomic mass × Given mass

⇒ 3g × 32g

= 96 g

weight of 3 grams of atoms of sulphur is 96 g.

Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 30
Save

16 gram of oxygen is equal to :-
Detailed Solution for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science - Question 30

To determine the equivalent quantities of oxygen, we can use the concept of atomic mass and Avogadro's number.
1. Atomic mass of oxygen: The atomic mass of oxygen is 16 grams per mole.
2. Avogadro's number: Avogadro's number is approximately 6.022 x 10^23 particles per mole.
Now, let's calculate the different quantities of oxygen:
A. 1 gram atom:
- The atomic mass of oxygen is 16 grams per mole.
- Therefore, 16 grams of oxygen is equal to 1 mole of oxygen.
- Since 1 mole of any element contains Avogadro's number of particles, 16 grams of oxygen is equal to 6.022 x 10^23 oxygen atoms.
- Hence, 16 grams of oxygen is equal to 1 gram atom.
B. 0.5 gram mole:
- A mole is a unit used to measure the amount of a substance.
- The atomic mass of oxygen is 16 grams per mole.
- Hence, 0.5 mole of oxygen is equal to (0.5 x 16) = 8 grams of oxygen.
- Therefore, 0.5 gram mole of oxygen is equal to 8 grams of oxygen.
C. 2 gram equivalents:
- An equivalent is a measure of the reactive capacity of a substance in a chemical reaction.
- For oxygen, the equivalent weight is equal to its atomic mass divided by its valence.
- The valence of oxygen is 2 since it can form two bonds with other elements.
- Therefore, the equivalent weight of oxygen is (16/2) = 8 grams per equivalent.
- Hence, 2 gram equivalents of oxygen is equal to (2 x 8) = 16 grams of oxygen.
D. All of these:
- Based on the calculations above, we can see that 16 grams of oxygen is equal to 1 gram atom, 0.5 gram mole, and 2 gram equivalents.
- Therefore, the answer is D, all of these.
In conclusion, 16 grams of oxygen is equal to 1 gram atom, 0.5 gram mole, and 2 gram equivalents.
View more questions
Information about Atoms And Molecules - Olympiad Level MCQ, Class 9 Science Page
In this test you can find the Exam questions for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science solved & explained in the simplest way possible. Besides giving Questions and answers for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science , EduRev gives you an ample number of Online tests for practice

Top Courses for Class 9

Download as PDF

Top Courses for Class 9

Important Questions for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science

Find all the important questions for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science at EduRev.Get fully prepared for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science with EduRev's comprehensive question bank and test resources. Our platform offers a diverse range of question papers covering various topics within the Atoms And Molecules - Olympiad Level MCQ, Class 9 Science syllabus. Whether you need to review specific subjects or assess your overall readiness, EduRev has you covered. The questions are designed to challenge you and help you gain confidence in tackling the actual exam. Maximize your chances of success by utilizing EduRev's extensive collection of Atoms And Molecules - Olympiad Level MCQ, Class 9 Science resources.

Atoms And Molecules - Olympiad Level MCQ, Class 9 Science

Prepare for the Atoms And Molecules - Olympiad Level MCQ, Class 9 Science within the Class 9 exam with comprehensive MCQs and answers at EduRev. Our platform offers a wide range of practice papers, question papers, and mock tests to familiarize you with the exam pattern and syllabus. Access the best books, study materials, and notes curated by toppers to enhance your preparation. Stay updated with the exam date and receive expert preparation tips and paper analysis. Visit EduRev's official website today and access a wealth of videos and coaching resources to excel in your exam.

Online Tests for Atoms And Molecules - Olympiad Level MCQ, Class 9 Science

Practice with a wide array of question papers that follow the exam pattern and syllabus. Our platform offers a user-friendly interface, allowing you to track your progress and identify areas for improvement. Access detailed solutions and explanations for each test to enhance your understanding of concepts. With EduRev's Online Tests, you can build confidence, boost your performance, and ace Atoms And Molecules - Olympiad Level MCQ, Class 9 Science with ease. Join thousands of successful students who have benefited from our trusted online resources.
© EduRev
Education Revolution
Signup to see your scores go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup