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If water samples are taken from the sea, rivers, clouds, lakes or snow, they will be found to contain hydrogen and oxygen in the ration of $1 : 8$. This clearly demonstrates the law of ?

evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

  1. Chemical equilibrium

  2. Definite proportions

  3. Combining volumes

  4. The relation of elements in the periodic tables

Show answer
Correct Option: B
Explanation:
In chemistry, the law of definite proportion states that a given chemical compound always contains its component elements in a fixed ratio (by mass) and does not depend on its source and method of preparation. Since in all the water samples given, taken from the sea, rivers, clouds, lakes or snow, the hydrogen and oxygen ratio is constant, hence it clearly demonstrates the law of definite proportions.

If you put a waterproof, nine-volt battery in a mug of water containing two teaspoonfuls of salt, the water will be broken into gases and bubbles will start forming. What are these gases?

evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

  1. Carbon dioxide and nitrogen

  2. Hydrogen and oxygen

  3. Fluorine and chlorine

  4. Sulphur dioxide and 'laughing gas'

Show answer
Correct Option: B
Explanation:

Water will start comparing electrostatically,


$2H _2O \underset{electroysis}{\rightarrow} 2H _2+O _2$


These gases are hydrogen and oxygen.

A sample of pure water, irrespective of source, contains 88.89% oxygen and 11.11% hydrogen
by mass. The data supports the:

evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

  1. law of conservation of mass

  2. law of constant composition

  3. law of multiple proportion

  4. law of reciprocal proportion

Show answer
Correct Option: B
Explanation:

According to the law of constant composition in a given compound, the elements always combine in the same proportion as each other.

Water($H _{2}O$) molecule consists of two hydrogen atoms of relative mass 1 and one oxygen atom of relative mass 16. 

Therefore, the ratio in which 2 atoms of hydrogen and one atom of oxygen are combined is $1:8$ and will always be the same irrespective of the source of water.

Therefore, the correct option is $B$

When one mole each of CO and $O _2$ are made to react at STP, the total number of moles at an end of the reaction is:

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  1. 1.5 moles

  2. 1 mole

  3. 4 moles

  4. 2 moles

Show answer
Correct Option: A
Explanation:

$2CO + O _2 \rightarrow 2CO _2$


Initially, we have one mole of each CO and $O _2$.

1 mole of CO will react with 0.5 mole of $O _2$ to form 1 mole of $CO _2$.

the total no. of moles at the end of the reaction is 0.5 mole $O _2$ + 1 mole of CO$ _2$= 1.5 moles

64g of an organic compound has 24 g carbon and rest hydrogen and oxygen.  The empirical formula of the compound is

evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

  1. ${C _2}{H _8}{O _2}$

  2. ${C _2}{H _4}O$

  3. $C{H _4}O$

  4. $C{H _2}O$

Show answer
Correct Option: C

Zinc sulphate contains $22.65$% $Zn$ and $43.9$% ${H} _{2}O$. If the law of constant proportions is true, then the mass of zinc required to give $40g$ crystal will be:

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  1. $9.06g$

  2. $90.6g$

  3. $0.906g$

  4. $906g$

Show answer
Correct Option: A
Explanation:
100g of crystals are obtained from = 22.65g of $Zn$

$\therefore $ 40 g of crystals are obtained from =$\frac { 22.65 }{ 100 } \times40$

= 9.06g 

Which of the following statements is/are correct?

evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

  1. A sample of $CaCO _3$ contains $Ca = 40\%,\ C = 12\%$ and $O= 48\%$. If the law of constant composition is true, then the mass of $Ca$ in $10$ g of $CaCO _3$ from another source is $4.0$ g.

  2. $12$ g of carbon is heated in vacuum and there is no change in the mass. This is the best example of the law of conservation of mass.

  3. Air is heated at constant pressure and there is no change in mass but the volume increases. This is the best example of the law of conservation of mass.

  4. $SO _2$ gas was prepared by (i) heating $Cu $ with conc. $H _2SO _4$ (ii) burning sulphur in oxygen (iii) reacting sodium sulphite $(Na _2SO _3)$ with dilute $H _2SO _4$. It was observed that is each case, $S$ and $O$ combine in the ratio of $1:1$. This data illustrates the law of constant composition.

Show answer
Correct Option: A,D
Explanation:

(A) A sample of $CaCO _3$ contains $Ca = 40\%$, $C = 12\%$ and $O= 48\%$. If the law of constant composition is true, then the mass of $Ca$ in $10$ g of $CaCO _3$ from another source is $\dfrac {40 \times 10 }{100}= 4.0 $ g


According to the law of constant composition, all samples of a given chemical compound have the same elemental composition by mass.
Hence, the statement A is true.

(B) $12$ g of carbon is heated in vacuum and there is no change in the mass. This is not the best example of the law of conservation of mass as there is no chemical transformation involved since there is a vacuum.
The Law of Conservation of Mass states that matter can be changed from one form into another, mixtures can be separated or made and pure substances can be decomposed but the total amount of mass remains constant.
Hence, the statement B is false.

(C) Air is heated at constant pressure and there is no change in mass but the volume increases. This is not the best example of the law of conservation of mass as there is no chemical transformation involved since there is a vacuum.

The Law of Conservation of Mass states that matter can be changed from one form into another, mixtures can be separated or made and pure substances can be decomposed but the total amount of mass remains constant.
Hence, the statement C is false.

(D) $SO _2$ gas was prepared by (i) heating $Cu$ with conc. $H _2SO _4$ (ii) burning sulphur in oxygen (iii) reacting sodium sulphite $(Na _2SO _3)$ with dilute $H _2SO _4$. It was observed that is each case, S and O combine in the ratio of $1:1$. This data illustrates the law of constant composition.

According to the law of constant composition, all samples of a given chemical compound have the same elemental composition by mass.

Hence, the statement D is true.

$14$ g of element X combines with $16$ g of oxygen. On the basis of this information, which of the following is a correct statement?

evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

  1. The element X could have an atomic weight of $7$ amu and its oxide is $XO$

  2. The element X could have an atomic weight of $14$ amu and its oxide is $\displaystyle X _{2}O$

  3. The element X could have an atomic weight of $7$ amu and its oxide is $\displaystyle X _{2}O$

  4. The element X could have an atomic weight of $14$ amu and its oxide is $\displaystyle XO _{2}$

Show answer
Correct Option: C
Explanation:

$14$ g of an element X combines with $16$ g of oxygen, then the element X could have an atomic weight of $7$ amu and its oxide is $X _2O $.
When the atomic weight is $7$, $14$ g will corresponds to $2$ moles.
$16$ g of oxygen corresponds to $1$ mole. 

The formula $X _2O $ suggests the valency of $1$ for X.

The % loss in mass after heating a pure sample of potassium chlorate (Mol. mass = 122.5) will be:

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  1. 12.25

  2. 24.50

  3. 39.17

  4. 49.0

Show answer
Correct Option: C
Explanation:

$2KClO _3\rightarrow 2KCl+3O _2$
$2\times 122.5$ grams shows a wieght loss of  $3\times32 grams$
So 245 grams of $KClO _3$ is 100%
96 grams of $O _2$ is X%
$x=\frac{100\times 96}{245}=39.17$.
Hence option C is correct.

The volume of oxygen at NTP evolved when 1.70 g of sodium nitrate is heated to a constant mass is: 

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  1. 0.112 litre

  2. 0224 litre

  3. 22.4 litre

  4. 11.2 litre

Show answer
Correct Option: B
Explanation:
The chemical equation is:

$2NaNO _3(s)→  2NaNO _2 (s)+ O _2(g)$

2 moles of NaNO$ _3$ give one mole of Oxygen/22.4 L oxygen at STP

weight of NaNO$ _3$ = 85g

2 moles of NaNO$ _3$ weigh = 2(23+14+48) = 170g

170 g of NaNO$ _3$ will produce 22.4 L Oxygen at STP

​1.70 g will produce $= \cfrac{1.70}{170}×22.4 =0.224$ litres

Hence, the correct option is $\text{B}$