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Multiple choice law of reciprocal proportion laws of chemical combination basic concepts of chemistry some basic concepts of chemistry chemistry

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

Multiple choice law of reciprocal proportion laws of chemical combination basic concepts of chemistry some basic concepts of chemistry chemistry

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 } $

Multiple choice law of reciprocal proportion laws of chemical combination basic concepts of chemistry some basic concepts of chemistry chemistry

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. $M _1O _3$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 ${ M } _{ 2 }{ O } _{ 3 }$
Multiple choice law of reciprocal proportion laws of chemical combination basic concepts of chemistry some basic concepts of chemistry chemistry

$CO _2$ 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. $3:2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In CO2, C is 27.27%, so O is 72.73%. In CS2, C is 15.79%, so S is 84.21%. In SO2, S is 50%, so O is 50%. The ratio of S:O in SO2 is 50:50, which simplifies to 1:1.

Multiple choice chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula water of crystallisation

What changes does not occur in presence of organic solvent when acidified $K _{2}Cr _{2}O _{7}$ reacts with $H _{2}O _{2}$ solutions

  1. Orange colour of solution turns blue
  2. $O.S$ if $Cr$ atom decreases
  3. $O.S$ of $Cr$ atom remains constant

  4. Unpaired electron remain same
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula water of crystallisation

From $\ 2 \ mg$ calcium $1.2\times 10^{19}$ atoms are removed. The number of $g$ - atoms of calcium left is $(Ca=40)$:

  1. $5\times 10^{-5}$
  2. $2\times 10^{-5}$
  3. $3\times 10^{-5}$
  4. $5\times 10^{-6}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$2\ mg$ Calcium moles $=\dfrac{2\times 10^{-3}}{40}=5\times 10^{-5}$


$1$ mole $=6.022\times 10^{23}$ atoms

$x=\dfrac{1.2\times 10^{19}}{6.022\times 10^{23}}$

$=1.99\times 10^{-5}$ moles removed
$\therefore$ Remaining moles $=5\times 10^{-5}-1.99\times 10^{-5}$
$=3.00\times 10^{-5}$ moles
$1\ g$ atom $=1$ mole
$\therefore$ Option $C$ correct.

Multiple choice chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula water of crystallisation

Find the mass percentages (mass %) of Na, H, C, and O in sodium hydrogen carbonate.

  1. 30, 20, 45, 5

  2. 28, 1, 14, 57

  3. 24, 23, 12, 1

  4. None of above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

22.99 g (1 mol) of Na
12.01          (1 mol) of H
48.0 (1 mol) of C
48.0 (3 mole $\times$ 16.00 gram per mole) of O
The mass of one mole of $NaHCO _3$ is $ 22.99 g + 1.01 g + 12.01 g + 48.00 g = 84.01 g$
And the mass percentages of the elements are
mass % $Na = \dfrac{22.99 g}{84.01 g} \times 100 = 27.36$%
mass % $H = \dfrac{1.01 g}{84.01 g} \times 100 = 1.20$%
mass % $C = \dfrac{12.01 g}{84.01 g} \times 100 = 14.30$%
mass % $O = \dfrac{48.00 g}{84.01 g} \times 100 = 57.14$%

Multiple choice chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula water of crystallisation

Determine the percentage composition of $K$ in $KMnO _4$.

  1. $31\%$
  2. $43.58\%$
  3. $25\%$
  4. $55\%$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
To find:- $\%$ composition of K in $KMnO _4$
A/c
Molar mass of $K=39$g
Molar mass of $KMnO _4=158$gm
$\%$ of k in $KMnO _4=\dfrac{39}{158}\times 100=24.68\%$
$\approx 25\%$.
Option C is correct.