Tag: law of constant proportion

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Multiple choice evs let's play with water law of constant proportion - i law of constant proportion 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:

  1. multiple proportions

  2. definite proportions

  3. reciprocal proportions

  4. conservation of mass
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion 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]

  1. $PbCl _3$
  2. $PbCl _2$
  3. $PbCl _4$
  4. $PbCl$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $PbCl _4$. 

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

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.

  1. Law of reciprocal proportion

  2. Law of multiple proportion

  3. Law of constant composition

  4. None of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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 :

  1. $0.016$ g
  2. $0.16$ g
  3. $1.6$ g
  4. $16$ g
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $C$

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

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?

  1. Law of constant proportions

  2. Law of multiple proportions

  3. Law of conservation of mass

  4. Law of constant volumes
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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 :

  1. Constant proportion

  2. Multiple proportion

  3. Reciprocal proportion

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

$1.375$ 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.

  1. Law of reciprocal proportion

  2. Law of multiple proportion

  3. Law of constant composition

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  1. Law of conservation of mass

  2. Law of constant composition

  3. Law of multiple proportion

  4. Law of reciprocal proportion
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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 :

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.