Tag: law of multiple proportion

Questions Related to law of multiple proportion

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


Correct Option: B
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.

Different proportions of oxygen in the various oxides of nitrogen prove the law of :

  1. equivalent proportion

  2. multiple proportion

  3. constant proportion

  4. conservation of matter


Correct Option: B
Explanation:

Different proportions of oxygen in the various oxides of nitrogen prove the law of multiple proportions.
According to the law of multiple proportions,
 "if two elements chemically combine with each other forming two or more compounds with different compositions
by mass then the ratios of masses of two interacting elements in the two compounds are small whole numbers".

Example.
14 g of nitrogen combine with 16  g of oxygen to form 30 g of $NO$.
The ratio of masses $N:O$ is $14:16$.

14 g of nitrogen combine with 32  g of oxygen to form 46 g of $NO _2$.
The ratio of masses $N:O$ is $14:32$.

The two ratios are in the proportion of $32:16=2:1$.

$3.2$ 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


Correct Option: C
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.

Law of multiple proportions is illustrated by which of the following pairs of compounds?

  1. $HCl$ and $HNO _{3}$

  2. $KOH$ and $KCl$

  3. $N _{2}O$ and $NO$

  4. $H _{2}S$ and $SO _{2}$


Correct Option: C
Explanation:

A fixed mass of nitrogen, say $100$ grams, may react with $57.14$ grams of oxygen to produce one oxide, or with $114.28$ grams of oxygen to produce the other. The ratio of the masses of oxygen that can react with $100$ grams of nitrogen is $57.14:114.28$ or  $2:1$, a ratio of small whole numbers.

Hydrogen peroxide and water contain $5.93$% and $11.2$% of hydrogen respectively. The data illustrates the law of:

  1. constant proportions

  2. multiple proportions

  3. reciprocal proportions

  4. conservation of mass


Correct Option: B
Explanation:

For hydrogen peroxide, 100 g of sample will contain 5.93 g hydrogen and  $\displaystyle 100 - 5.93 = 94.07$ g oxygen respectively.

The ratio of the mass of oxygen to the mass of hydrogen in hydrogen peroxide is  $\displaystyle \dfrac {94.07}{5.93} = 15.86$

For water, 100 g of sample will contain 11.2 g hydrogen and  $\displaystyle 100 - 11.2 = 88.8$ g oxygen respectively.

The ratio of the mass of oxygen to the mass of hydrogen in hydrogen peroxide is  $\displaystyle \dfrac {88.8}{11.2} = 7.93$

The two ratios are in the proportion $\displaystyle \dfrac {15.86}{7.93} = 2:1$

Hence, this illustrates the Law of multiple proportions. According to this law, if two elements chemically combine with each other forming two or more compounds with different compositions by mass then the ratios of masses of two interacting elements in the two compounds are small whole numbers.

Two different oxides of manganese are compare in the table shown below.

Color % Ms (by mass) % O (by mass)
Oxide # $1$Oxide # $2$ BlackDk Green $63.19$$77.50$ $36.81$$22.50$

When substituted X and Y in the fraction below, which pair CORRECTLY gives a result that illustrates the Law of Multiple Proportions?
$\underline {63.19}$
$X$
$\overline {77.50}$
$\overline {Y}$

  1. $X = 54.94$

    $Y = 54.94$

  2. $X = 36.81$

    $Y = 22.50$

  3. $X = 16.00$

    $Y = 16.00$

  4. $X = 22.50$

    $Y = 36.81$


Correct Option: A

The molecules of nitrogen monoxide and nitrogen dioxide differ by a multiple of the mass of one oxygen. The statement can be understood by the concept of :

  1. Law of multiple proportion

  2. Nuclear fusion

  3. Van dar Waals forces

  4. Graham's Law of Diffusion(Effusion)

  5. Triple point


Correct Option: A
Explanation:

The molecules of nitrogen monoxide and nitrogen dioxide differ by a multiple of the mass of one oxygen. The statement can be understood by the concept of the Law of multiple proportion.

Which of the following pairs of compounds can be used to illustrate Dalton's Law of Multiple Proportions?

  1. $NH _{4}$ and $NH _{4}Cl$

  2. $SO _{2}$ and $SO _{3}$

  3. $H _{2}$ and $O _{2}$

  4. $H _{2}O$ and $HCl$


Correct Option: B
Explanation:

Law of Multiple proportions states that if two elements form more than one compound between them, then the ratios of masses of second element which combine with a fixed mass of first element ratios of small whole numbers.

$*$ Ex: 1) $CO, CO _2$           2) $SO _2, SO _3$           3)$CuO, Cu _2O$ etc.

$CaCO _3 + 2HCl \rightarrow CaCl _2 + H _2O + CO _2$

The mass of calcium chloride formed when 2.5 g of calcium carbonate is dissolved in excess of hydrochloric acid is:

  1. 1.39 g

  2. 2.78 g

  3. 5.18 g

  4. 17.8 g


Correct Option: B
Explanation:

CaCO3 + 2HCl -----> CaCl2 + H2O + CO2
2.5g? g 
Mass of CaCO3= 40+12+3(16)=52+ 48= 100. 
Molar mass of CaCO3=100g. 
Mass of CaCl2= 40 + 2(35.5)= 40+ 71= 111. 
Molar mass of CaCl2=111g. 
Mass of CaCO3 (g) Mass of CaCl2 (g) 
For 100g of $CaCO _3$, 111g of CaCl_2$ is formed.
let for 2.5g of $CaCO_3$, $x$ g of $CaCl_2$ is formed.
Thus, by cross multiplication,

$x=111\times 2.5/100= 2.775g = 2.78g$.

For describing law of multiple proportions, which among the following pair can be used?

  1. NO and $NO _2$

  2. $CH _4$ and $CO _2$

  3. $ZnO _2$ and $ZnCl _2$

  4. $NH _3$ and $NH _4Cl$

  5. $H _2O$ and HCl


Correct Option: A
Explanation:

For describing law of multiple proportions, the pair $\displaystyle  NO$ and $\displaystyle  NO _2$ can be used.
According to the law of multiple proportions, if two elements combine to form more than one compound, then the different masses of one element which combine with fixed mass of other element, bear a simple ratio to one another.
In the pair $\displaystyle  NO$ and $\displaystyle  NO _2$, we have two elements N and O that combine to form more than one compound. In all other examples, more than two elements are present.