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 basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

The mass of nitrogen per gram in hydrazine is exactly one and half the mass of nitrogen in the compound ammonia. 


The fact illustrates the:

  1. law of conservation of mass

  2. multiple valency of nitrogen

  3. law of multiple proportion

  4. law of definite proportion
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The law of multiple proportions says that, If two elements form more than one compound between them, then the ratios of the masses of the second element which combine with a fixed mass of the first element will be ratios of small whole numbers.


The ratio of masses of nitrogen per gram of hydrogen in Hydrazine and Ammonia is 1.5:1 = 3:2 Which is simple whole-number ratio.

Therefore, it represents the law of multiple proportions.

Option C is correct.

Multiple choice chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

1 g of an oxide of A contained 0.5 g of A, 4 g of another oxide of A contained 1.6 g of A. These figures illustrate the law of:

  1. conservation of mass

  2. reciprocal proportions

  3. multiple proportions

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

According to the law of multiple proportion, when two elements combine with each other to form more than one compound, the weights of one element that combines with fixed weight of the other are in the ratio of small whole numbers.

Multiple choice chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

$i$. Percentage of $Mg$ in $MgO$ and $MgCl _{2}$ 

$ii$. Percentage of $C$ in $CO$ and $CO _{2}$
$iii$. Percentage of $Cr$ in $K _{2} Cr _{2}O _{7}$ and $K _{2}CrO _{4}$ 
$iv$. Percentage of $Cu$ isotopes in $Cu$ metal
The law of multiple proportions may be illustrated by data given by:

  1. $ii$
  2. $iv$
  3. $i$ and $ii $
  4. $i,\ ii$ and $iii$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

According to law of multiple proportion, if two elements combine to form more than one compound, then the weight of one element combining with the fixed weight of other, in the two compounds, is in a whole number ratio.

Multiple choice chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

The composition (by atoms) of compound A is $40$% X and $60$% Y. The composition (by atoms) of compound B is $25$% X and $75$% Y. According to the law of multiple proportions, the ratio of the weight of element Y in compounds A and B is:

  1. $1 : 2$
  2. $2 : 1$
  3. $2 : 3$
  4. $3 : 4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The composition of compound A is  $X:Y = 40:60= 2:3$
The composition of compound B is $X:Y= 25:75= 1:3$

According to the law of multiple proportions, 
In compound A $2$ moles of $X$ bind with $3$ moles of $Y$.
In compound B $2$ moles of $X$ bind with 6 moles of $Y$ 

The ratio of the weight of element $Y$, in combining with a fixed number of $X$ atoms, in these two compounds $A$ and $B$ is $ 3:6 = 1 : 2$.
Multiple choice chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

In $SO _{2}$ and $SO _{3}$, the ratio of weight of oxygen that combines with a fixed weight of sulphur is $2 : 3$. This illustrates the law of:

  1. constant proportions

  2. conservation of mass

  3. multiple proportions

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

The law of multiple proportions says that If two elements form more than one compounds between them, then the ratios of the masses of the second element which combine with a fixed mass of the first element will be ratios of small whole numbers.

Multiple choice chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

Elements A and B combine to form three different compounds:
$0.3$ g of A $+ 0.4$ g of B $\rightarrow$ $0.7$ g of compound X
$18.0$ g of A $+\ 48.0$ g of B $\rightarrow$ $66.0$ g of compound Y
$40.0$ g of A $+\ 159.99$ g of B $\rightarrow$ $199.99$ g of compound Z
State the law illustrated by these chemical combinations.

  1. Law of reciprocal proportion

  2. Law of multiple proportion

  3. Law of constant composition

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

Weights of B that combine with $1.0$ g of A in the compounds X, Y and Z are, respectively,
$\dfrac{0.4}{0.3}=1.33,\ \dfrac{48.0}{18.0}=2.66$ and $\dfrac{159.99}{40.0}=4.00$
Ratio being $1.33:2.66:4.00$ or $1:2:3$ is the simple ratio and this illustrates the law of multiple proportions.

Multiple choice chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

$3.2$ g sulphur combines with $3.2$ g of oxygen to form a compound in one set of conditions. In another set of conditions, $0.8$ g of sulphur combines with $1.2$ g of oxygen to form another compound. State the law illustrated by these chemical combinations.

  1. Law of constant composition

  2. Law of reciprocal proportion

  3. Law of multiple proportion

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

First case,
$3.2$ g of S combines with $3.2$ g of $O _2$.
$1$ g of S combines with $1$ g of $O _2$.
Second case,
$0.8$ g of S combines with $1.2$ g of $O _2$
$1$ g of S combines with $\dfrac {1.2}{0.8}=1.5$ g of $O _2$
Thus, the ratio of the $O _2$ in both cases which combines with a fixed mass ($1$ g) of $S=1:1.5$ or $2:3$, which is a simple whole number ratio and hence, the law of multiple proportion is verified.

Multiple choice chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

A metal forms two oxides with first oxide having 1 and second having 3 oxygen atoms and their masses are 74g and 164 g respectively. If the gram atomic mass of oxygen is 16g, do both compounds follow the law of multiple proportions?

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The law of multiple proportions states that if two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers. Since both compounds are oxides of the same metal, they follow this law.

Multiple choice chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

The chloride of a solid metallic element contains 57.89% by mass of the element. The specific heat of the element is $0.0324\, cal\, deg^{-1}\, g^{-1}$. Calculate the exact atomic mass of the clement. 

  1. 195.2

  2. 195

  3. 195.9

  4. 194
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
% of element is given in this question = 57.89
so %  of chlorine is = 100-57.89 = 42.11

Equivalent wt of metal= $\dfrac{57.89}{42.11} \times 35.5$ = 48.8

specific heat of metal= 0.0324

$\rightarrow $ According to Dulong and Petit's law
Approx atomic wt of metal= $\dfrac{6.4}{ specific heat} = \dfrac{6.4}{0.0324}$ = 197.53

Valency of metal= Approximate atomic wt/ Equivalent wt

 = $\dfrac{197.53}{ 48.8}$ = 4.04  = 4

Exact atomic wt of metal = $ Equivalent wt of metal\times Valency of metal$ = 48.8 x 4
  = 195.2
answer is A
Multiple choice chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

In compound A, 1.0 g nitrogen combines with 0.57 g oxygen. In compound B, 2.0 g nitrogen unite with 2.24 g oxygen and in compound C, 3.0 g nitrogen combine with 5.11 g oxygen. These results obey the law of: 

  1. multiple proportions

  2. constant proportions

  3. reciprocal proportions

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

In compound A 1 gm of nitrogen combined with 0.57 gm of oxygen and compound B 1 gm of nitrogen combined with 1.12 gm of oxygen and C 1 gm of nitrogen combined with 1.7 gm of oxygen so the fixed amount of nitrogen is 1: 2 :3 so this follows of multiple proportion rule 

Multiple choice chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

Carbon and oxygen form two compounds. Carbon content in one of them is $42.9$% and in the other is $27.3$%. The given data is in agreement with:

  1. law of conservation of mass

  2. law of multiple proportions

  3. law of reciprocal proportions

  4. law of definite proportions
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The given data is in agreement with Law of multiple proportions.

The Law of multiple proportions states that the masses of one element combine with a fixed mass of the second element are in a ratio of whole numbers.
In $100$ $g$ of the first compound $O:C=$$57$ $/42.9$ $=1.33$  
In $100$ $g$ of the second compound $O:C=$$72.7$ $/27.3$ $=2.66$  
Ratio $=2.66:1.33=2:1$
This whole number ratio is consistent with Law of multiple proportions. 

Multiple choice chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

$12g$ of carbon combines with $64g$ sulphur to form ${CS} _{2}$. $12g$ carbon also combines with $32g$ oxygen to form ${CO} _{2}$. $10g$ sulphur combines with $10g$ oxygen to form ${SO} _{2}$. These data illustrate the

  1. Law of multiple proportions

  2. Law of definite proportions

  3. Law of reciprocal proportions

  4. Law of gaseous volumes
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given,

$12g$ $C$ combines with $64g$ $S$ to give $CS _2$
$12g$ $C$ combines with $32g$ $O$ to give $CO _2$
Also, $10gS$ combines with $10gO$ to give $SO _2$
In $CS _2$ $C:S=12:64=3:16$
In $CO _2$ $C:O=12:32=3:8$
From, $CS _2$ & $CO _2$, $S:O=16:8=2:1$
Also in $SO _2$ $S:O=10:10=1:1$
$\Rightarrow 2:1$ is multiple of $1:1$
$\cfrac {1}{1}\times 2=\cfrac {2}{1}$
$\Rightarrow $ Law of reciprocal proportions is illustrated here.

Multiple choice chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

Carbon and oxygen from two compounds. Carbon content in one of them is 42.9% and in the other is 27.3%. The given data is in agreement with :

  1. law of conservation of mass

  2. law of multiple proportions

  3. law of reciprocal proportions

  4. law of definite proportions.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since the two compounds are composed of the same two elements (carbon and oxygen) with different mass percentages, they illustrate the law of multiple proportions.

Multiple choice chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

$P _xO _y$ what will be correct value of $x$ and $y$ if $P$ and $H$ combine in the mass ratio of $3.1 : 0.3$ and in water $H$ and $O$ combine in the mass ratio $0.2 : 1.6$?

  1. $1$ and $1$
  2. $1$ and $2$
  3. $2$ and $3$
  4. $2$ and $5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

 

$P:H = 3.1 : 0.3 = (3.1 \times : 0.30) \times 2 = 6.2 : 0.6$

$H:O = 0.2 : 1.6 = (0.2 \times 3 : 1.6) \times 3 = 0.6 : 4.8$

$P:O = 6.2 : 4.8 = 3.1 : 2.4 $ (ratio by mass) $= 1.29: 1$

$1.29$ mass $P= 1 $ mass $O$

$x$ mass $P= 16 \ g$ mass $O$

$x= 16 \times 1.29 = 20.64 \ g$ Phosphorous

$= \cfrac {20.64}{31}$ moles Phosphorous

$= 0.66$

$1 \ mole \ O(y) = 0.66 \ mole \ P(x) = \cfrac 23$

$\therefore \ x:y = 2:3 $

$x =2 \ ; \ y=3$

Multiple choice chemistry hard water and soft water heavy water study of heavy water hydrogen and its compounds

The weight of one mole of heavy water is:

  1. $18$ gm
  2. $20$ gm
  3. $19$ gm
  4. $21$ gm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
 Deuterium is called as heavy water,

As atomic mass of Deuterium is 2 and atomic mass of oxygen is 16

So Molar mass of $D _2O$ is = $2\times2 +1\times 16= 20$

Weight of $1$ mole of ${ D } _{ 2 }O =$ Molecular mass of ${ D } _{ 2 }O = 20$ gm

Option B is correct.