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 quantitative chemistry avogadro hypothesis avogadro's law avogadro law

What is the mass of $6.022\times { 10 }^{ 23 }$ formula units of ${({NH} _{4})} _{2}{SO} _{4}$?

  1. $234.11\ g$
  2. $132.11\ g$
  3. $210.29\ g$
  4. $342.14\ g$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$1\space mole$ of $(NH _4) _2SO _4$ has $6.023 \times 10^{23}$ formula units.

So, the mass of $6.023 \times 10^{23}$ formula units of $(NH _4) _2SO _4$. is molar mass of $(NH _4) _2SO _4$

Molar mass of $(NH _4) _2SO _4$ $= 2(14 + 4) + 32 + 4\times 16 = 132 \space g$

Multiple choice chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

One mole of any substance contains $6.022\times 10^{23}$ atoms/molecules. Number of molecules of $H 2SO _4$ present in $100$mL of $0.02$M $H _2SO _4$ solution is _________?

  1. $12.044\times 10^{20}$ molecules
  2. $6.022\times 10^{23}$ molecules
  3. $1\times 10^{23}$ molecules
  4. $12.044\times 10^{23}$ molecules
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Moles of $H _2SO _4=$ Molarity of $H _2SO _4\times $ Volume of solution $(L)$


$=0.02\times 0.1$


$=2\times 10^{-3}$ moles

No. of $H _2SO _4$ molecules $=2\times 10^{-3}\times 6.022\times 10^{23}=12.044\times 10^{20}$ molecules


Therefore, the correct option is A.

Multiple choice chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

The number of gram molecule of oxygen in $6.02 \times 10^{24}$ CO molecules is :

  1. 10 gm molecules

  2. 5 gm molecules

  3. 1 gm molecules

  4. 0.5 gm molecules
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given :


$6.02\times 10^{24}$ $CO$ molecules


We know,

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

$6.023\times 10^{23}$ molecules of $CO$ contain $6.023\times 10^{23}$ atoms of $O$

Then, $6.02\times 10^{24}$ $CO$ molecules contain $6.02\times 10^{24}$ atoms of $O$

$\Rightarrow$ no. of gram atoms of $O=\cfrac {6.02\times 10^{24}}{6.02\times 10^{23}}$

$=10gm$ atoms  $\longrightarrow 1$

As oxygen is a diatomic molecule.

No. of gm of molecules of oxygen=$\cfrac {10gm\quad atoms}{2\quad atoms/molecule}=5gm$ molecules

Therefore, the correct option is B.

Multiple choice chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

The number of g-atom of oxygen in $6.02\times { 10 }^{ 24 }$ $CO$ molecule is

  1. $1$
  2. $0.5$
  3. $5$
  4. $10$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

One molecule of $CO$ contains one oxygen atom
$\therefore 6.02\times { 10 }^{ 24 }\quad $ molecules of $CO$ contain $6.02\times { 10 }^{ 24 }$ oxygen atoms
$6.02\times { 10 }^{ 23 }$ atoms of oxygen $\equiv$ $1g$ atom of oxygen
$\therefore 6.02\times { 10 }^{ 24 }$ atoms of oxygen $\equiv$ $10g$ atom of oxygen.

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

The charge required to deposit 40.5 g of Al(atomic mass=27.0 g) from the fused ${ Al } _{ 2 }{ \left( { SO } _{ 4 } \right) } _{ 3 }$ is :

  1. $4.34\times { 10 }^{ 5 }C$
  2. $43.4\times { 10 }^{ 5 }C$
  3. $1.44\times { 10 }^{ 5 }C$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The reaction is Al^3+ + 3e- -> Al. 1 mole of Al (27g) requires 3 Faradays (3 * 96500 C). For 40.5g, the moles of Al = 40.5 / 27 = 1.5 moles. Charge = 1.5 * 3 * 96500 = 4.5 * 96500 = 434250 C, which is 4.34 * 10^5 C.

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

Which of the following solutions contains the greatest number of ions assuming that each solute is fully ionized?

  1. $1$ cm$^{-3}$ of $0.2M$ ${Na} _{2}{SO} _{4}$
  2. $1$ cm$^{-3}$ of $0.2M$ $Ca{Cl} _{2}$
  3. $1$ cm$^{-3}$ of $0.2M$ $NaCl$
  4. $1$ cm$^{-3}$ of $0.2M$ ${Cr} _{2}{({SO} _{4})} _{3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

No. of ions in 11 molecule of $Na _2SO _4=3$ 

No. of ions in 11 molecule of $CaCl _2=3$
No. of ions in 11 molecule of $NaCl=2$
No. of ions in 11 molecule of $Cr _2(SO _4) _3=5$
As the molarity and volume is same for all the given substances, number of ions will be greatest for the compound having greatest number of ions in one molecule. Therefore, $Cr _2(SO _4) _3$ will have the greatest number of ions.

Multiple choice imperfections in solids solid state the solid state chemistry

The composition of a sample of iron oxide is $Fe _{0.93}O _{1.0}$. What percentage of iron is present in the form of Fe(III)?

  1. $5.053$
  2. $10.053$
  3. $15.053$

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

Let the percentage amount of Fe(III) be x and the amount of Fe(II) be $1-x$. Hence,  
$3x+2(1-x) = 2/0.93 = 2.15053$
On solving, we get, $x = 0.15053$
So, % of Fe(III) = 15.053

Multiple choice biology food and it's constituents energy giving food carbohydrate and its test biological molecules

The ratio between hydrogen and oxygen in carbohydrate is

  1. $5 : 1$
  2. $4 : 3$
  3. $3 : 1$
  4. $2 : 1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Monosaccharide is the simplest form of carbohydrate which is composed of carbon, hydrogen and oxygen. The basic formula of the monosaccharide is C$ _n$H$ _{2n}$O$ _n$. These are considered as simplest form of sugar. The ratio between the hydrogen and oxygen is 2:1. For example, fructose is a monosaccharide with five carbon atom, ten hydrogen and five oxygen.
So, the correct answer is option D.
Multiple choice chemistry principles of metallurgy concentration of ore concentration of ores general principles of metallurgy

In the equation,


$4M + 8 CN^- + 2 H _2O + O _2 \longrightarrow 4[M(CN) _2]^- + 4 OH^-$

the metal M is:

  1. Copper

  2. Silver

  3. Gold

  4. Zinc
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

Cyanide process, also called Macarthur-Forrest Process, method of extracting silver and gold from their ores by dissolving them in a dilute solution of sodium cyanide or potassium cyanide. 


The process was invented in 1887 by the Scottish chemists John S. MacArthur, Robert W. Forrest, and William Forrest.

$4Ag + 8 CN^- + 2 H _2O + O _2 \longrightarrow 4[Ag(CN) _2]^- + 4 OH^-$


$4Au + 8 CN^- + 2 H _2O + O _2 \longrightarrow 4[Au(CN) _2]^- + 4 OH^-$


Hence options B & C are correct.

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

What is the density of an iron block which has a volume of ${12 cm^3}$ and a mass of 96g?

  1. ${ 8 kg/m ^3}$
  2. ${80 kg /m ^3}$
  3. ${8000 kg /m ^3}$
  4. ${800 kg/ m ^3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Answer is A.

As we know, Density $d = \dfrac{Mass, M}{ Volume, V}$


In this case, an iron block which has a volume of ${12 cm^3}$ = ${0.12 m^3}$ and a mass of $96 g = 0.096 kg$.

Therefore, Density $d =\dfrac{ 0.096}{0.12} = 8 kg/m^{ 3 }$.

Hence, the density of the iron block is 8 $kg/m^{ 3 }$.

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

Equal volumes of iron and aluminum have different masses because they have different _________ .

  1. shapes

  2. densities

  3. lengths

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

If volume of iron = Volume of aluminium

then mass of iron > mass of aluminium
$\rho _{iron}\times V _{iron} > \rho _{aluminium}\times V _{aluminium}$
$\rho _{iron} > \rho _{aluminium}$

Multiple choice chemistry chemistry of non-metals oxides of elements simple oxides oxygen

A metal forms two oxides. The higher oxide contains $80\%$ metal. $0.72 g$ of the lower oxide gave $0.8 g$ of higher oxide when oxidized. Then the ratio of the weight of oxygen that combines with the fixed weight of the metal in the two oxides will be:

  1. $2 :3$
  2. $2: 1$
  3. $4 :5$
  4. $3 : 2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given that ,in higher oxide 80% metal is true.
wt. of metal$=\dfrac { 80 }{ 100 } \times 0.8\\$       
                    $ =0.64gm\\ $
wt. of oxygen$=0.16gm$
in lower oxide,
wt. of metal will be same as that of higher oxidation state(+2)

wt. of oxygen$=(72-0.64)=0.08g$

so,ratio of weight of $O _2$ & meals in two oxide is $0.16:0.08=2:1$.
1000ml of 0.75 M HCl = 0.75 mol

So, 25 ml of HCl will contain HCl = $0.75 \times 25 /1000$  = 0.01875

2 mol of HCl reacts with 1 mol of $CaCO _3$

So, 0.01875 mol of HCl will react with $\frac{1}{2}$ $\times$ 0.01875  = $0.009375$ mol

Molar mass of $CaCO _3$ = 100 g

Hence, the mass of 0.009375 mol of $CaCO _3$  = no. of moles $\times $ molar mass$
                                                                          
 $0.009375$ $\times100$  
                                                                   
Hence, the correct option is $C$.
Multiple choice chemistry chemistry of non-metals oxides of elements simple oxides oxygen

Two oxides of a metal contain $50$% and $40$% metal (M) respectively. If the formula of first oxide is $MO _2$, the formula of second oxide will be:

  1. $MO _2$
  2. $MO _3$
  3. $M _2O$
  4. $M _2O _5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\text{Compound } 1\quad\quad\quad\quad 50\% M+50\% O$
$\text{formula}\quad\quad\quad\quad MO _2$
Let atomic weight of metal be $X,$
$\Longrightarrow MO _2:$    $\cfrac{16}{X}=\cfrac{1}{2}$
Hence atomic weight of metal $=32$
$\text{Compound }2:$ $40\% M=\cfrac{40}{32}=1.25=1\\60\%O=\cfrac{60}{16}=3.75=3$
$\therefore$ Formula is $MO _3$.
Multiple choice chemistry chemistry of non-metals oxides of elements simple oxides oxygen

Two oxides of a metal contain 72.4% and 70% of metal respectively? If formula of 2nd oxide is $ M _{2}O _{3} $, find that of the first ?

  1. MO
  2. $ M _{3}O _{4} $
  3. $ M _{3}O _{2} $
  4. $ MO _{2} $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For the second oxide M2O3, 70% is metal, so 30% is oxygen. The ratio of moles M:O is (70/M) : (30/16) = 2:3. Solving gives M = 56 (Iron). For the first oxide (72.4% metal), ratio is (72.4/56) : (27.6/16) = 1.29 : 1.725 = 3:4. Thus, M3O4.