Chemistry

Mole Concept and Stoichiometry

365 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 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

If the % composition of Cl in HCl is 46%, what is the mass of Cl in HCl?

  1. 13.3 g

  2. 66 g

  3. 25 g

  4. 16.76 g
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\%$ $ composition =\cfrac{Mass \quad of\quad an \quad element.}{Total \quad mass \quad of\quad compound.}$

$Total \quad mass\quad of\quad HCl=1+35.5 g=36.5g$
$\therefore$ $\cfrac{46}{100}=\cfrac{mass\quad of\quad Cl}{36.5}\ mass\quad of\quad Cl=16.76g$

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

Oxalic acid dihydrate, $H _{2}C _{2}O _{4}\cdot 2H _{2}O(s)$ is often used as a primary reagent to standardise sodium hydroxide solution. Which of these facts are reasons to choose this substance as a primary standard?
I. It is diprotic.
II. It is a stable compound that can be weighed directly in air.
III. It is available in pure form.

  1. III only

  2. I and II only

  3. II and III only

  4. I, II and III
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l} { H _{ 2 } }{ C _{ 2 } }{ O _{ 4 } }\cdot 2{ H _{ 2 } }O\left( s \right) \to oxadic\, acid\, dihydrate\,  \ Since,\, it\, is\, very\, stable\, and\, also\, available\, in\, pressure \ from\, in\, the\, laboratory\, i.e.\, it's\, always\, used\, as\, primary\, s\tan  dard\, in\, the\, lab. \ Hence,\, the\, option\, C\, is\, the\, correct\, answer. \end{array}$

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.