Tag: basic concepts of chemistry

Questions Related to basic concepts of chemistry

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:

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

  1. $1 : 2$

  2. $2 : 1$

  3. $2 : 3$

  4. $3 : 4$

Show answer
Correct Option: A
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$.

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:

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

  1. constant proportions

  2. conservation of mass

  3. multiple proportions

  4. reciprocal proportions

Show answer
Correct Option: C
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.

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.

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

  1. Law of reciprocal proportion

  2. Law of multiple proportion

  3. Law of constant composition

  4. None of the above

Show answer
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 :

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

  1. equivalent proportion

  2. multiple proportion

  3. constant proportion

  4. conservation of matter

Show answer
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.

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

  1. Law of constant composition

  2. Law of reciprocal proportion

  3. Law of multiple proportion

  4. None of the above

Show answer
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?

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

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

  2. $KOH$ and $KCl$

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

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

Show answer
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:

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

  1. constant proportions

  2. multiple proportions

  3. reciprocal proportions

  4. conservation of mass

Show answer
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?
$\underline {63.19}$
$X$
$\overline {77.50}$
$\overline {Y}$

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

  1. $X = 54.94$

    $Y = 54.94$

  2. $X = 36.81$

    $Y = 22.50$

  3. $X = 16.00$

    $Y = 16.00$

  4. $X = 22.50$

    $Y = 36.81$

Show answer
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 :

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

  1. Law of multiple proportion

  2. Nuclear fusion

  3. Van dar Waals forces

  4. Graham's Law of Diffusion(Effusion)

  5. Triple point

Show answer
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?

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

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

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

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

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

Show answer
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.