Tag: let's play with water

Questions Related to let's play with water

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


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?

  1. Law of constant proportions

  2. Law of multiple proportions

  3. Law of conservation of mass

  4. Law of constant volumes


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 :

  1. Constant proportion

  2. Multiple proportion

  3. Reciprocal proportion

  4. None of these


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.

  1. Law of reciprocal proportion

  2. Law of multiple proportion

  3. Law of constant composition

  4. None of these


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:

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

  2. $NaOH$ and $KOH$

  3. $NaCl$ and $NaBr$

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


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:

  1. Law of conservation of mass

  2. Law of constant composition

  3. Law of multiple proportion

  4. Law of reciprocal proportion


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 :

  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


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.

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 :

  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


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 :

  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


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?

  1. $12\ g$

  2. $13\ g$

  3. $11\ g$

  4. $10\ g$


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.