Tag: law of definite proportions

Questions Related to law of definite proportions

$2.16$ grams of Cu, on reaction with $HNO _{3}$, followed by ignition of the nitrate, gave $2.7$ g of copper oxide. In another experiment $1.15$ g of copper oxide, upon reaction with hydrogen, gave $0.92$ g of copper. This data illustrate the law of:

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  1. multiple proportions

  2. definite proportions

  3. reciprocal proportions

  4. conservation of mass

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Correct Option: B
Explanation:

In the first sample of copper oxide (obtained by the action of nitric acid on Cu), the mass of Cu is 2.16 g and the mass of oxygen is $\displaystyle 2.7 - 2.16  =  0.54$ g. 

The ratio of the mass of Cu to the mass of oxygen is $\displaystyle \dfrac {2.16}{0.54} = 4 :1$

In the second sample of copper oxide (which reacts with hydrogen) , the mass of Cu is 0.92 g and the mass of oxygen is $\displaystyle 1.15 - 0.92  =  0.23$ g. 
The ratio of the mass of Cu to the mass of oxygen is $\displaystyle \dfrac {0.92}{0.23} = 4 :1$
Hence, this illustrates the law of definite Proportions.

An experiment showed that a lead chloride solution is formed when 6.21 g of lead combines with 4.26 g of chlorine. What is the empirical formula of this chloride? 

[Pb = 207; Cl = 35.5]

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  1. $PbCl _3$

  2. $PbCl _2$

  3. $PbCl _4$

  4. $PbCl$

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Correct Option: C
Explanation:
  Mass        Atomic weight    Relative no. of moles    Simplest ratio
Lead           6.21 g        207     6.21/207 = 0.03    0.03/0.03 = 1
Chlorine 4.26 g        35.5     4.26/35.5 = 0.12    0.12/0.03 = 4


Hence, empirical formula is $PbCl _4$. 

Common salt obtained from Clifton beach contained $60.75\%$ chlorine while $6.40$ g of a sample of common salt from Khewra mine contained $3.888$ g of chlorine. State the law illustrated by these chemical combinations.

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  1. Law of reciprocal proportion

  2. Law of multiple proportion

  3. Law of constant composition

  4. None of the above

Show answer
Correct Option: C
Explanation:

First case :


Common salt from Clifton beach contains $=$ $60.75\%$ $Cl _2$

$100$ g of salt $=60.75$ g of $Cl _2$

$1$ g of salt $=\dfrac {60.75}{100}=0.6075$ g of $Cl _2$

Second case :


$6.40$ g of $NaCl$ from Khewra mine $=3.888$ g of $Cl _2$

$1$ g of $NaCl$ from Khewra mine $=\dfrac {3.888}{6.40}=0.6075$ g of $Cl _2$

Thus, the weight of $Cl _2$ in $1$ g of salt in both the cases is same. Hence, the law of constant composition is verified.


Hence the correct option is C.

A sample of calcium carbonate $\displaystyle \left ( CaCO _{3} \right )$ has the percentage composition as given: $Ca = 40\%,\ C = 12\%,\ O = 48\%$. 


If the law of constant proportions is true, then the weight of calcium in $4$ g of a sample of calcium carbonate obtained from another source will be :

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  1. $0.016$ g

  2. $0.16$ g

  3. $1.6$ g

  4. $16$ g

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Correct Option: C
Explanation:

In $100 $ g $\displaystyle CaCO _{3}$, weight of $Ca$ is $40 $ g.

In $4$ g $\displaystyle CaCO _{3}$, weight of $Ca$ is $=\displaystyle \frac{40}{100}\times 4=1.6$ g

Hence, the correct option is $C$

1.2375 g of cupric oxide on being heated in a current of hydrogen gave 0.9322 g of the metal In another experiment 0.9369 g of pure copper was dissolved in nitric acid Excess of acid evaporated and the residue has ignited The weight of the cupric oxide left was 1.2469 g. Which law of chemical combination is shown by the above results?

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  1. Law of constant proportions

  2. Law of multiple proportions

  3. Law of conservation of mass

  4. Law of constant volumes

Show answer
Correct Option: B
Explanation:

law of multiple proportion is shown as copper reacts with two different compounds .

The percentage of copper and oxygen in a sample of CuO obtained from different methods were found to be same. This proves the law of :

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  1. Constant proportion

  2. Multiple proportion

  3. Reciprocal proportion

  4. None of these

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Correct Option: A
Explanation:

The percentage of copper and oxygen in a sample of CuO obtained from different methods were found to be same. This proves the law of Constant proportion as the ratio of Cu:O remains constant

$1.375$ g of cupric oxide was reduced by heating in a current of hydrogen and the weight of copper obtained was $1.098$ g. In another experiment, $1.156$ g of copper was dissolved in nitric acid and the resulting solution was evaporated to dryness. The residue of copper nitrate when strongly heated was converted into $1.4476$ g of cupric oxide. State the law illustrated by these chemical combinations.

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  1. Law of reciprocal proportion

  2. Law of multiple proportion

  3. Law of constant composition

  4. None of these

Show answer
Correct Option: C
Explanation:

Law of constant composition states that all samples of a given chemical compound have the same elemental composition by mass. 

Here, in the first experiment 1.375 g of cupric oxide was reduced by heating in a current of hydrogen and the weight of copper obtained was 1.098 g.
Which means 1.375 g of cupric oxide has 1.098 g of carbon and $1.375-1.098 = 0.277$ g of oxygen. 
Copper and oxygen are in the ration $\frac{1.098}{0.277} = \frac{1}{4}$. 
In the second experiment 1.156 g of copper was dissolved in nitric acid to form copper nitrate, which on strongly heating got converted into 1.4476 g of cupric oxide. 
Which means 1.4476 g of cupric oxide has 1.156 g of copper and $ 1.4476-1.156 =0.2911 g $ oxygen.
Copper and oxygen are in the ratio $\frac{1.156}{0.2911} = \frac{1}{4}$.
Which mean what ever may be the source of a compound, it has same elemental composition by mass in all samples.

A pair of compounds which is an illustration of law of multiple proportions is:

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  1. ${H} _{2}O$ and ${D} _{2}O$

  2. $NaOH$ and $KOH$

  3. $NaCl$ and $NaBr$

  4. $CO$ and ${CO} _{2}$

Show answer
Correct Option: D
Explanation:

Carbon combines with oxygen in $1:1$ ratio to form carbon monoxide and $1:2$ ratio to form carbon dioxide.

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

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  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:

This law states that all samples of a given chemical compound have the same elemental composition by mass e.g. oxygen makes 8/9th of the mass of pure water while hydrogen makes up the remaining 1/9th of the mass.

Hydrogen combines with nitrogen in a 3:14 weight ratio to form ammonia. If every molecule of ammonia contains three atoms of hydrogen and one atom of nitrogen, an atom of nitrogen must weigh :

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  1. 14 times the mass of a hydrogen atom

  2. 14/3 times the mass of a hydrogen atom

  3. 3 times the mass of a hydrogen atom

  4. 3/14 times the mass of a hydrogen atom

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Correct Option: A
Explanation:

Atoms of different elements combine in simple whole number ratios to form chemical compounds. Combination always takes place in the simplest possible way between particles of different weights.