Tag: law of constant proportion - i

Questions Related to law of constant proportion - i

Hydrogen combines with oxygen in a 1:8 weight ratio to form water. If every molecule of water contains two atoms of hydrogen and one atom of oxygen, an atom of oxygen must weigh :

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  1. 8 times the mass of a hydrogen atom

  2. 16 times the mass of a hydrogen atom

  3. 1/16 times the mass of a hydrogen atom

  4. 1/8 times the mass of a hydrogen atom

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Correct Option: B
Explanation:

The law of definite proportions and constant composition states that all samples of a given chemical compound have the same elemental composition by mass e.g. oxygen makes up up about 8/9th of mass of any sample of pure water while hydrogen makes up the remaining 1/9th of mass.

When elements react, their atoms combine in :

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  1. a simple whole number ratio (unique for each pair of elements)

  2. arbitrary proportions

  3. simple whole number ratios (more than one possible)

  4. 1:1 ratios

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Correct Option: C
Explanation:

The law of multiple proportions state 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.

When $3\ g$ of carbon is burnt in $8\ g$ of oxygen, $11 g$ of carbon dioxide is produced. What mass of carbon dioxide will be formed when $3\ g$ of carbon is burnt in $50\ g$ oxygen?

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  1. $12\ g$

  2. $13\ g$

  3. $11\ g$

  4. $10\ g$

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Correct Option: C
Explanation:

This is governed by the law of definite proportion. Carbon and oxygen combine in $3:8$ ratio. So $11 \ g$ of carbon dioxide is formed. 


Hence, in the latter case too, only $8 \ g$ of $O _2$ will be reacted with 3g of carbon (because carbon is the limiting reagent here ) to form 11g of $CO _2$ and  $42 \ g$ of oxygen will remain unreacted.

Hence, the correct option is C.

When $1.375\ g$ of cupric oxide is reduced on heating in a current of hydrogen, the weight of copper remaining $1.098\ g$. In another experiment, $1.179\ g$ of copper is dissolved in nitric acid and resulting copper nitrate converted into cupric oxide by ignition. The weight of cupric oxide formed is $1.476 \ g$. This is in agreement with :

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  1. Law of definite proportion

  2. Law of mass conservation

  3. Law of momentum conservation

  4. None of the above

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Correct Option: A
Explanation:

First experiment
$\bullet$ Copper oxide $=1.375 g$
$\bullet$ Copper left $= 1.098 g$
$\bullet$ Oxygen present $= 1.375 - 1.098 = 0.277 g$
Percentage of oxygen in $CuO = (0.277) (100$ %) $1.375 = 20.15$ %

Second Experiment
$\bullet$ Copper taken $= 1.179 g$
$\bullet$ Copper oxide formed $= 1.476 g$
$\bullet$ Oxygen present $= 1.476 - 1.179 = 0.297 g$
Percentage of oxygen in $CuO = (0.297) (100$ %) $1.476 = 20.12$ %
Percentage of oxygen is approximately (within significant figures limit) the same in both the above cases. So the law of constant composition is illustrated.

In chemistry, the law of definite proportion, sometimes called Proust's law or the law of definite composition, or law of constant composition states that a given chemical compound always contains its component elements in fixed ratio and does not depend on its source and method of preparation.

Who stated the law of constant proportion?

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  1. Archimedes

  2. Dalton

  3. Proust

  4. Antony Lavoisier

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Correct Option: C
Explanation:

The law ofconstant proportion is given by Proust.

A 78.0 g sample of an unknown compound contains 12.4 g of hydrogen.  What is the percent by mass of hydrogen in the compound?

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  1. 25%

  2. 35%

  3. 16%

  4. 50%

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Correct Option: C
Explanation:

% of $H$$=$$\dfrac{12.4}{78}\times100$$=16$%

The percentage of copper and oxygen in samples of CuO obtained by different methods were found to be the same. This illustrates the law of:

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  1. Constant proportion

  2. Conservation of mass

  3. Multiple proportion

  4. Reciprocal proportion

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Correct Option: A
Explanation:

In chemistry, the law of definite proportions, sometimes called Proust's law or the law of definite composition, states that a chemical compound always contains exactly the same proportion of elements by mass and this law further says that, in any particular chemical compound, all samples of that compound will be made up of the same elements in the same proportion or ratio. For example, any water molecule is always made up of two hydrogen atoms and one oxygen atom in a ratio.

Which of the following statement is incorrect for the composition of water?

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  1. For water, the gravimetric or mass ratio of hydrogen to oxygen is $1:8$

  2. In water, the ratio of hydrogen and oxygen is $2:1$ by volume

  3. The ratio in which hydrogen and oxygen combine gives the composition of water.

  4. None of the above

Show answer
Correct Option: D
Explanation:

All of the following statements are correct for the composition of water.
A
For water, the gravimetric or mass ratio of hydrogen to oxygen is 1:8.
B
In water, the ratio of hydrogen and oxygen is 2:1 by volume.
C
The ratio in which hydrogen and oxygen combine gives the composition of water.
Note:
$2H _2 + O _2 \rightarrow 2H _2O$
Thus, the composition of water is $H : O = 2: 1 $
It is the ratio in which hydrogen and oxygen combine. The atomic masses of H and O are 1 and 16 respectively. Hence, for water, the gravimetric or mass ratio of hydrogen to oxygen is 2:16 or 1:8.
In water, the ratio of hydrogen and oxygen is 2:1
by volume. It is equal to the ratio in which hydrogen and oxygen combine.

Nitrogen reaction with oxygen to form nitrogen (III) oxide.


What will be the volume ratio of the reactants and products of the given reaction?

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  1. 2 : 3 : 2

  2. 2 : 3 : 4

  3. 1 : 3 : 2

  4. 2 : 1 : 2

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Correct Option: A
Explanation:

$2 N _2 + 3 O _2 \rightarrow 2N _2 O _3$

reactants = $N _2 \, \& \, O _2$
Product = $N _2O _3$
Ratio are $2 : 3 : 2$ by law of constant proportions 
{stionchiometry coefficient) 


A sample of $CaCO _3$ has Ca - 40%, C = 12% and 0 = 48%. If the law of constant proportions is true, then the mass of Ca in 5 g of $CaCO _3$ from another source will be:

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  1. 2.0 g

  2. 0.2 g

  3. 0.02 g

  4. 20.0 g

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Correct Option: A
Explanation:

Since, mass percentage of Ca in $CaCO _3 = 40 \text{%} $ and law of constant proportion is true.
So, the mass percentage of $Ca $ in $ 5gm$ $CaCO _3$ will also be $40\text%.$
Hence,$ 40\text{% of 5gm =}$$ 5\times $$\dfrac{40}{100}$ $= 2gm $
Hence, answer is option A.