Tag: laws of chemical combination

Quizzes (0) Questions (76)

Questions Related to laws of chemical combination

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $32:16=2:1$.

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

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 $SO _{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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. $X = 22.50$

    $Y = 36.81$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To illustrate the Law of Multiple Proportions with mass percentages, one element's mass is held constant while the ratio of the masses of the other element in the two oxides is compared. Using the oxygen percentages directly as X = 36.81 and Y = 22.50 satisfies this ratio when the mass of manganese is scaled or compared.

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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 $HCl$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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,

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