Tag: basic concepts of chemistry

Questions Related to basic concepts of chemistry

"Law of Reciprocal Proportion" is also known as:

  1. Law of Equivalent Proportion

  2. Law of Multiple Proportions

  3. Law of Permanent Ratios

  4. Law of Proportion


Correct Option: A,C
Explanation:

Law of Equivalent Proportion and Law of Permanent Ratios are the other names for the Law of Reciprocal Proportion.

Two volumes of ammonia, on dissociation gave one volume of nitrogen and three volumes of hydrogen. How much hydrogen will be obtained from dissociation of $40 \,mL\, of\, NH _3$ ? 

  1. 60

  2. 40

  3. 80

  4. 50


Correct Option: A
Explanation:

Its easy as $NH _3 : N _2 : H _2$ IS 2:1:3 As $NH _3$ is 40 ml hence form 60 ml of h2 and 20 ml of n2

so the answer is A

Which one of the following sets of compounds correctly illustrate the law of reciprocal proportions.

  1. $P _{2}O _{3}, PH _{3}, H _{2}O$

  2. $P _{2}O _{5}, PH _{3}, H _{2}O$

  3. $N _{2}O _{5}, NH _{3}, H _{2}O$

  4. $N _{2}O, NH _{3}, H _{2}O$

  5. $NO _{2}, NH _{3}, H _{2}O$


Correct Option: A
Explanation:

Law of Reciprocal proportions:-

When two different elements combine with the same mass of a $3^{rd}$ element, the ratio of masses in which they do so must be same or multiple of the mass ratio in which they combine with each other.
$A)$ $P _2O _3,PH,H _2O$
The ratio of phosphorous combining with oxygen in $P _2O _3$ is 
$P:O=62:48=31:24$
The ratio of Phosphorous combining with hydrogen in $PH _3$ is
$P:H=31:3$
The ratio of hydrogen combining with oxygen in $H _2O$ is 
$H:O=2:16=1:8$       $\longrightarrow 1$
Now, the ratio of hydrogen & oxygen combining with $P$
$\Rightarrow 3:24=1:8$     $\longrightarrow 2$
From $1$ & $2$,  $1:8$ is same in both cases
$\therefore$ Law of reciprocal proportions is illustrated correctly.

Law of reciprocal proportions is also called :

  1. The law of equivalent proportions

  2. Law of permanent ratios

  3. Law of gaseous volume

  4. All of the above


Correct Option: A,B
Explanation:

Law of reciprocal proportions is also called as the law of equivalent proportions and law of permanent ratios.

$A$ g of sulphur and $ B$ g of oxygen are allowed to react to form $SO _2 $. What will be the ratio of $ A:B$ ?

  1. $1:1$

  2. $1:2$

  3. $1:3$

  4. $1:4$


Correct Option: A
Explanation:
$S\left( 32ℊ \right) +{ O } _{ 2 }\left( \left( 16\times 2 \right) ℊ \right) \longrightarrow S{ O } _{ 2 }$
$\Rightarrow $Ratio of A and B is $1:1$

7.95 gm oxide of copper on reaction with hydrogen shows a mass loss and the residue obtained weighs 6.35 gm. In another experiment, 19.05 gm of copper was dissolved in nitric acid and the copper nitrate produced was converted to oxide weighing 23.85 gm. This data illustrates:

  1. Law of multiple proportion

  2. Law of mass conservation

  3. Law of reciprocal proportion

  4. Law of constant proportion


Correct Option: D
Explanation:

The ratio of Masses of oxide of $Cu$ to $Cu$,

In experiment $A=\cfrac{7.95}{6.35}=1.25$
In experiment $B=\cfrac{23.85}{19.05}=1.22$
It shows the law of constant proportion.

The law of reciprocal proportions was proposed by :

  1. Jeremias Richter

  2. Joseph proust

  3. Joseph louis

  4. Antoine


Correct Option: A
Explanation:

Jeremias Richter (Jeremias Benjamin Richter) was a German chemist. He proposed the law of Reciprocal proportions in $1792$.

Which of the following sets of compounds correctly illustrate the law of reciprocal proportion ?

  1. $N _2O,NH _3,SO _2$

  2. $P _2O _5,PH _3,H _2O$

  3. $NO _2,NH _3,SO _3$

  4. None of these


Correct Option: B
Explanation:

The law of reciprocal proportions is one of the basic laws of stoichiometry. It relates the proportions in which elements combine across a number of different elements. A simple statement of the law is:

 Element A combines with element B and also with C, then, if B and C combine together, the proportion by weight in which they do so will be simply related to the weights of B and C which separately combine with a constant weight of A.

Hydrogen combines with Oxygen to form ${ H } _{ 2 }O$

$2:16\\ 1:8$

Phosphorous combines with Oxygen to form ${ P } _{ 2 }O _{ 5 }$

$62:80\\ 31:40\\ $

Phosphorous combines with Hydrogen to form $P{ H } _{ 5 }$

$31:5\\ \cfrac { 31 }{ 40 } \times \cfrac { 8 }{ 1 } =\cfrac { 31 }{ 5 } $

Two oxides of metal contain $27.6$% and $30$% of oxygen respectively. if the first one is $M _3O _4$ then which of the following will be the other one?

  1. $M _3O _3$

  2. $M _2O _3$

  3. $M _4O _2$

  4. $M _1O _3$


Correct Option: B
Explanation:
Given, ${ M } _{ 3 }{ O } _{ 4 }$; Let mass of the metal$=x$
% of metal in ${ M } _{ 3 }{ O } _{ 4 }=\cfrac { 3x }{ 3x+64 } \times 100$
But as given % age$=\left( 100-27.6 \right) =72.4$
So, $\left( \cfrac { 3x }{ 3x+64 }  \right) \times 100=72.4$
$\Rightarrow x=56$
In 2nd oxide, oxygen$=30$%, So metal$=70$%
So, the ratio is $M:O$
$\cfrac { 70 }{ 56 } :\cfrac { 30 }{ 16 } $
$1.25:1.875$
$2:3\Rightarrow $ oxide is ${ M } _{ 2 }{ O } _{ 3 }$

$CO _2$ contains $27.27$% of carbon,$CS _2$ contains $15.79$% of carbon,$SO _2$ contains $50$% sulphur. What will be the ratio of $S:O$ in $SO _2$?

  1. $1:1$

  2. $1:2$

  3. $2:1$

  4. $3:2$


Correct Option: A