Chemistry

Mole Concept and Stoichiometry

354 Questions

The mole concept is a core area of chemistry focusing on the quantitative relationships between reactants and products in chemical reactions. Questions cover molecular mass calculations, stoichiometric conversions, and determining the number of molecules. This topic is essential for most science entrance and competitive examinations.

molecular mass calculationsstoichiometric conversionsmole fraction problemsheavy water reactionsoxidation calculations

Mole Concept and Stoichiometry Questions

Multiple choice law of reciprocal proportion laws of chemical combination basic concepts of chemistry some basic concepts of chemistry chemistry

$CO _2$ contains $27.27$% of carbon,$CS _2$ contains $15.79$% of carbon,$SO _2$ contains $50$% sulphur. What will be the ratio of $S:O$ in $SO _2$?

  1. $1:1$
  2. $1:2$
  3. $2:1$
  4. $3:2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In CO2, C is 27.27%, so O is 72.73%. In CS2, C is 15.79%, so S is 84.21%. In SO2, S is 50%, so O is 50%. The ratio of S:O in SO2 is 50:50, which simplifies to 1:1.

Multiple choice chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula water of crystallisation

Find the mass percentages (mass %) of Na, H, C, and O in sodium hydrogen carbonate.

  1. 30, 20, 45, 5

  2. 28, 1, 14, 57

  3. 24, 23, 12, 1

  4. None of above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

22.99 g (1 mol) of Na
12.01          (1 mol) of H
48.0 (1 mol) of C
48.0 (3 mole $\times$ 16.00 gram per mole) of O
The mass of one mole of $NaHCO _3$ is $ 22.99 g + 1.01 g + 12.01 g + 48.00 g = 84.01 g$
And the mass percentages of the elements are
mass % $Na = \dfrac{22.99 g}{84.01 g} \times 100 = 27.36$%
mass % $H = \dfrac{1.01 g}{84.01 g} \times 100 = 1.20$%
mass % $C = \dfrac{12.01 g}{84.01 g} \times 100 = 14.30$%
mass % $O = \dfrac{48.00 g}{84.01 g} \times 100 = 57.14$%

Multiple choice chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula water of crystallisation

Given:
Mass of beaker is X g
Mass of beaker + mixture is Y g
Mass of washed and dried sand is Z g
Find the percentage of sand in the mixture.

  1. $\frac{Y-X}{100}\times Z$
  2. $\frac{Y-X}{Z}\times 100$
  3. $\frac{Z}{Y-X}\times 100$
  4. $\frac{100}{Y-X}\times Z$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Mass of mixture=$Y-X$

Percentage of sand= $\frac{weight   of   sand}{weight   of   mixture}\times 100$
                                 =$\frac{Z}{Y-X}\times100$

Multiple choice chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula water of crystallisation

The mass of a sand and powdered mixture along with a beaker is 56 g. If the mass of the dried mixture is 20 g, find the % composition of the mixture in 100 g?
(weight of beaker = 20 g).

  1. 20%

  2. 36%

  3. 55%

  4. 60%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Mass of beaker = 20 g
Mass of mixture + beaker = 56 g
Mass of mixture = 56 - 20 = 36 g
Mass of washed and dried sand = 20 g
100 g of mixture contains $= \dfrac{20}{36} \times 100 = 55$% of sand

Multiple choice chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula water of crystallisation

Calculate the % composition of Carbon in $CO _2$. Molar mass is 44.01.

  1. 40%

  2. 67%

  3. 30%

  4. 27%
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Molar mass of compound:
Mass due to carbon:   12.01 g/mol
Total molar mass $= 12.01 + 2(16.00) = 44.01g/mol$             ($CO _2$ has two $O _2$ atoms)
Percent composition of carbon: $12.01/44.01 \times 100 = 27.28$%

Multiple choice chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula water of crystallisation

Mass of beaker=20 g
Mass of beaker+mixture=50 g
Mass of washed and dried sand=5 g
Find the percentage of sand in the mixture.

  1. 16.67%

  2. 20%

  3. 25%

  4. 30%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Mass of mixture=$50 g-20 g=30 g$

Mass of washed and dried sand in the mixture=$5g$
Percentage=$\frac{5}{30}\times 100$
                   =16.67%


Multiple choice chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula water of crystallisation

Mass of beaker=10g
Mass of mixture+beaker=25g
Mixture contains 30% sand. Find the weight of sand.

  1. 5 g

  2. 4.5 g

  3. 3 g

  4. 3.5 g
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Mass of mixture=$25g-10g=15g$

Percentage of sand=30%=$\frac{Mass  of  sand}{Mass  of  mixture}$
 Mass of sand =$0.3\times15g=4.5 g$

Multiple choice chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula

The number of $g$ molecules of oxygen in $6.023\times 10^{24} \ CO$ molecules is ________________.

  1. $1\ g$ molecule
  2. $0.5\ g$ molecule
  3. $5\ g$ molecules
  4. $10\ g$ molecules
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$1$ mole of $CO=6.023\times 10^{23}$ molecules

$x$ mole of $CO=6.023\times 10^{24}$ molecules
$\therefore x=10$ moles $=10\ g$ molecules
$CO=1$ oxygen atom
$\therefore$ Oxygen molecules $=\dfrac{10}{2}$
$=5\ g$ molecules 
Option $C$ is correct.

Multiple choice chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula

If two compounds have the same empirical formula but different molecular formula, they must have :

  1. different percentage composition

  2. different molecular weights

  3. same vapour density

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If two compounds have the same empirical formula but different molecular formula, they must have different molecular weights.


For example, $CH _2O$ and $C _6H _{12}O _6$ have the same empirical formula but different molecular formula, they have different molecular weights.

Option B is correct.

Multiple choice chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula

Calculate the number of molecules present in $0.5$ moles of magnesium oxide $\left( MgO \right) $. 

[Atomic weights :  $Mg=24, O=16$]

  1. $14.09\times { 10 }^{ 23 }$ molecules
  2. $3.0115\times { 10 }^{ 23 }$ molecules
  3. $30.12\times { 10 }^{ 23 }$ molecules
  4. $14.09\times { 10 }^{ -23 }$ molecules
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
1 mole of MgO contains Avogadro's number of molecules which is $ \displaystyle 6.023 \times 10^{23}$ molecules.

0.5 mole of MgO will contain $ \displaystyle 0.5 \times 6.023 \times 10^{23}=3.0115 \times 10^{23}$ molecules.
Multiple choice zoology heredity and variation effects of smoking smoking tobacco drugs

Out of 4000 chemicals released by cigarette, potent carcinogens are

  1. 12

  2. 43

  3. 216

  4. 172
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A substance that causes cancer is called a carcinogen. Cigarette smoke contains more than 4,000 chemicals out of which 43 are carcinogenic. Some of the carcinogens are nicotine, formaldehyde, benzene, carbon monoxide, etc. Most of the cigarette smoke remains in the lungs. This smoke damages the lung walls causing cancer. Thus the correct answer is option B.

Multiple choice bio-chemistry biological oxidation classical idea of redox reactions oxidation reduction redox reaction

The mass of $H _{2}O _{2}$ that is completely oxidised by $30.2\ g$ of $KMnO _{4}$ (molar mass= $158\ g mol^{-1}$) in acidic medium is 

  1. $12\ g$
  2. $14\ g$
  3. $16\ g$
  4. $1\ g$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In acidic medium, 2 moles of KMnO4 react with 5 moles of H2O2. Calculating moles of KMnO4 (30.2g / 158g/mol = 0.191 mol), the required moles of H2O2 is (5/2) * 0.191 = 0.4775 mol. Mass = 0.4775 * 34g/mol = 16.2g, which rounds to 16g.