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.

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Mole Concept and Stoichiometry Questions

Multiple choice chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

The molecular mass of a salt of oxy acid of chlorine of a divalent metal which contains more number of oxygen atoms than its corresponding '-ic' acid is $239\ g/mole$.
What is the atomic mass of the metal?

  1. 24

  2. 39

  3. 40

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

The formula of the salt with molecular weight 239 is $Ca(ClO _{4}) _{2}$ and thus the divalent metal is calcium with atomic weight 40.

Multiple choice chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

How many gram-equivalents of NaOH are required to neutralise 25 cm$^3$ of a decinormal HCl solution ?

  1. 0.00125

  2. 0.0025

  3. 0.0050

  4. 0.025
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
For acid-base reactions,
$\underset{(HCl)}{g-equivalent of acid}\, =\, \underset{(NaOH)}{g-equivalent of base}$
$\therefore$ g-equivalent of acid = $(25\,\times\, 10^{-3})dm^3\, \times\, \displaystyle \frac{1}{10}$
$=\, 25\, \times\, 10^{-4}$
$=\, 0.0025$
Multiple choice chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

The number of gram-molecules of oxygen in $6.022\times 10^{24}$ molecules of $CO$ is:

  1. $10$ gm moles
  2. $5$ gm moles
  3. $1$ gm mole
  4. $0.5$ gm mole
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Number of oxygen atoms = Number of $CO$ molecules$=6.022\times 10^{24}$

Number of oxygen molecule $=\dfrac12\times$  Number of oxygen atoms $=3.011\times 10^{24}$
Number of g-molecule of $O _2$ molecules $=\dfrac{3.011\times 10^{24}}{6.022\times 10^{23}}=5\ gm\ mole$

Multiple choice chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

The molar mass of $CuSO _4.5H _2O$ is 249. Its equivalent mass in the reaction (a) and (b) would be:
(a) Reaction $CuSO _4 + KI \rightarrow$ product
(b) Electrolysis of $CuSO _4$ solution

  1. (a) 249 (b) 249

  2. (a) 124.5 (b) 124.5

  3. (a) 249 (b) 124.5

  4. (a) 124.5 (b) 249
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In water, ${ CuSO } _{ 4 }\cdot 5{ H } _{ 2 }O$ dissociates into ${ Cu }^{ 2+ }$ & ${ SO } _{ 4 }^{ 2- }$. The ions are doubly charged. Hence half as much is needed to react with a compound that dissociates into singly charged species.

(a) Reaction of ${ CuSO } _{ 4 }+KI\longrightarrow $ products.
     Eq. wt of ${ CuSO } _{ 4 }=249/2=124.5$
(b) Electrolysis of ${ CuSO } _{ 4 }$
     At Cathode ${ \underset { +2 }{ Cu }  }^{ 2+ }+{ 2e }^{ - }\longrightarrow \underset { 0 }{ Cu } $
$\therefore$   Charge in oxidation state $=2=n-$factor
$\therefore$   Eq. wt $=\dfrac { 249 }{ 2 } =124.5$

Multiple choice chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

10 ml of 0.1 M solution sodium hydroxide is completely neutralised by 25 ml of 3 gram of dibasic acid in one solution the molecular weight of acid is:

  1. 225 g

  2. 250 g

  3. 300 g

  4. 150 g
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given,
Concentration of $NaOH=0.1 N$
Volume$=10\,mL$
Mass of Dibasic acid$=3g$
Volume$=25\,mL$
Using,
$n _{1}M _{1}V _{1}=n _{2}M _{2}V _{2}$
$n _{2}=2$
$\Rightarrow 1\times 0.1\times 10=2\times25\times\cfrac{3}{M}$
$\Rightarrow M=\cfrac{2\times 25\times 3}{10\times 0.1}$
$\Rightarrow M=150\,g$
Multiple choice chemistry ionic equilibrium introduction to ionic equilibria in solution ionic equilibrium in solution ionisation of weak acids and weak bases

The value of observed and calculate molecular weights of silver nitrate are $92.64$ and $170$ respectively. 


The degree of dissociation of silver nitrate will be :

  1. $60\%$
  2. $83.5\%$
  3. $85.3\%$
  4. $41.6\%$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The degree of dissociation alpha is calculated by (normal molar mass - observed molar mass) / (observed molar mass * (n-1)). For AgNO3, n=2. (170-92.64) / (92.64 * 1) = 77.36 / 92.64 = 0.835, which is 83.5%.

Multiple choice chemistry mix and separate different types of solutions various mixtures introduction to solutions

$100\ ml$ of an aqueous solution contains $6.0\times {10}^{21}$ solute molecules. The solution is diluted to $1$ lit. The number of solute molecules present in $10\ ml$ of the dilute solution is:

  1. $6.0\times {10}^{20}$
  2. $6.0\times {10}^{19}$
  3. $6.0\times {10}^{18}$
  4. $6.0\times {10}^{17}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
100 ml solution diluted to 1 liters (1000) ml contains $6.0 \times 10^{21}$ solute molecular

No of molecules present in 10 ml

$ = \dfrac{10 \times 6.0 \times 10 ^{21}}{1000} = 6 \times 10 ^{19} $

Multiple choice chemistry mix and separate different types of solutions various mixtures introduction to solutions

When $1$ mole of a substance is present in $1$ L of the solution, it is known as :

  1. normal solution

  2. molar solution

  3. molal solution

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

Molar concentration is a measure of the concentration of a solute in a solution and its unit is mol L$^{-1}$. Molarity is a method to express the concentration of a solution. It is defined as the number of moles of solute dissolved per liter of solution. 


Hence, when $1$ mole of a substance is present in $1$ L of the solution, it is known as a molar solution.

Option B is correct.

Multiple choice chemistry mix and separate different types of solutions various mixtures introduction to solutions

10g of sodium hydroxide dissolved in 1 L of water to make _____ solution.

  1. $0.25 M$
  2. $0.5 M$
  3. $1 M$
  4. $1.5 M$
  5. $4 M$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
molar mass of sodium hydroxide $=$$40g/mole$
$Molarity = \dfrac{Mass\, of\, solute}{(Molar\, mass\, of\, the\, solute)\times(Volume\, of \, solution\, in\, litres)}$

So $Molarity$ $=$$10/(40\times1)$$=$$0.25M$
Multiple choice chemistry mix and separate different types of solutions various mixtures introduction to solutions

How much water, in liters, must be added to 0.5 L of 6 M HCl to make it 2 M?

  1. 0.33

  2. 0.5

  3. 1

  4. 1.5

  5. 2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Molarity =$\dfrac{Mass \,of\, the\, solute}{(Molar \,mass\, of\, the\, solute)\times {(Vol. of\, soln.\, in\, liters)}}$

Mass of solute will remain same before and after mixing water.
so 
       $M _1V _1$$=$$M _2V _2$
or 
       $V _2$$=$$6\times0.5/2$$=$$1.5L$
this is final total volume so water added $=$$1.5-0.5$$=$$1L$ 
Multiple choice chemistry atom fundamental particles of an atom the structure of atoms discovery of subatomic particles

Suppose the chemists hed selected $10^{20}$ as the number of particles in a mole. The molar mass of oxygen gas would be (Use Avogadro number $=6.0\times 10^{23}$)

  1. $5.33\times 10^{-3}g$
  2. $5.35\times 10^{-23}g$
  3. $5.33\times 10^{-43}g$
  4. $32\times 10^{3}g$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Molar mass of oxygen we know is $32$g.

$\therefore 1$ molecule oxygen weighs $\cfrac { 32 }{ 6\times { 10 }^{ 23 } } $g
$\Rightarrow { 10 }^{ 20 }$ molecules would weigh $\cfrac { 32 }{ 6\times { 10 }^{ 23 } } \times { 10 }^{ 20 }$
$\therefore 1 $mole would weigh $=5.33\times { 10 }^{ -3 }$g.

Multiple choice chemistry atom fundamental particles of an atom the structure of atoms discovery of subatomic particles

The number of electrons present in $100ml$ of $0.1N$ ${H} _{2}{SO} _{4}$ is:

  1. $6.85\times {10}^{22}$
  2. $7.50\times {10}^{23}$
  3. $1.5\times {10}^{23}$
  4. $1.8\times {10}^{22}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Normality=$n\times Molarity$

$\therefore 0.1=2\times \cfrac { moles\quad  of\quad  H _2SO _4}{ 0.1 }$
$\therefore Moles\quad  of\quad  H _2SO _4=0.005\quad moles$

$1$ molecule of $H _2SO _4$ contains $50$ electrons.

No. of molecules in $0.005$ moles=$0.005\times6.022\times {10}^{ 23 }$
=$0.03011\times {10}^{ 23 }$ molecules

$\therefore$ No. of electrons in $0.005$ moles=$0.03011\times {10}^{ 23 }\times 50$
=$1.5\times {10}^{ 23 }$ electrons

Multiple choice chemistry atom fundamental particles of an atom the structure of atoms discovery of subatomic particles

Total number of electrons present in $14\ gm$ of nitrogen gas is:

$[N _A = 6\times 10^{23}]$

  1. $6\times { 10 }^{ 23 }$
  2. $4.2\times { 10 }^{ 24 }$
  3. $3\times { 10 }^{ 23 }$
  4. $4.2\times { 10 }^{ 23 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

14g of N2 gas is 0.5 moles (molar mass 28g/mol). Each N2 molecule has 14 electrons (7 per nitrogen atom), so 0.5 moles of N2 contains 0.5 * 6 * 10^23 * 14 = 4.2 * 10^24 electrons.

Multiple choice chemistry atom fundamental particles of an atom the structure of atoms discovery of subatomic particles

What is the total number of electrons present in 1.6 g of methane?

  1. $6.023 \times 10^{23}$

  2. 16
  3. $12.04 \times 10^{23}$
  4. $6.023 \times 10^{24}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
1 mole methane = 16 g
Electrons in 1 mole of methane are $=6+4 \times 1=10$
Mole of methane in 1.6 g=$\cfrac{mass}{molar\ mass}$
$=\cfrac{1.6}{16}=0.1$ mole
$\therefore$ 0.1 mol methane contains $0.1 \times 10$mol electron=1 mole of electron
1 mole electron=$6.023 \times 10^{23}$ electrons