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Hydrogen combines with oxygen to form $H _2O$ in which 16 g of oxygen combine with 2 g of hydrogen. Hydrogen also combines with carbon to form $CH _4$ in which 2 g of hydrogen combine with 6 g of carbon.If carbon and oxygen combine together then they will do show in the ratio of 

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

  1. 13 : 32

  2. 6 : 16

  3. 1 : 2

  4. 12 : 24

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Correct Option: B

Two gases found dissolved in natural water are 

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

  1. oxygen and carbon dioxide

  2. hydrogen and oxygen

  3. sulphur dioxide and hydrogen

  4. chlorine and ammonia

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Correct Option: A

A sample of pure carbon dioxide, irrespective of its source contains 27.27 % carbon and 72.73% oxygen. The given data supports: 

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

  1. Law of constant composition

  2. Law of conservation of mass

  3. Law of reciprocal proporties

  4. Law of multiple proportions

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Correct Option: A

In Haber's process, the volume of ammonia relative to the total volume of reactants at STP is:

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

  1. one fourth

  2. one half

  3. same

  4. three fourth

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Correct Option: B
Explanation:

This reaction occurs in Haber's process-


$N _2$ + $3H _2$ $\rightarrow$ $2NH _3$

The total volume of reactants at STP is 4x

The volume of ammonia is 2x.

Hence, Volume of ammonia = $\dfrac{1}{2}$(Volume of reactant)

$2.16$ 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:

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

  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]

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

  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.

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

  1. Law of reciprocal proportion

  2. Law of multiple proportion

  3. Law of constant composition

  4. None of the above

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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 :

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

  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?

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

  1. Law of constant proportions

  2. Law of multiple proportions

  3. Law of conservation of mass

  4. Law of constant volumes

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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 :

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

  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