Tag: the language of chemistry

Questions Related to the language of chemistry

What is the atomic mass of sodium?

chemistry the language of chemistry atomic weight and molecular weight masses of atoms and molecules chemical analysis and formulae

  1. $22\ g/mol$

  2. $23\ g/mol$

  3. $11\ g/mol$

  4. $20\ g/mol$

Show answer
Correct Option: B
Explanation:

A relative atomic mass is a measure of how heavy atoms are. It is the ratio of the average mass per atom of an element from a given sample to 1/12 the mass of a carbon-12 atom.

The relative atomic mass in 1 mole of isotopes of sodium atoms is 23 g/mol.

What is the relative atomic mass of cadmium?

chemistry the language of chemistry atomic weight and molecular weight masses of atoms and molecules chemical analysis and formulae

  1. 112.411 amu

  2. 20.00 amu

  3. 21.00 amu

  4. 20.43 amu

Show answer
Correct Option: A
Explanation:

The relative atomic mass is an average of the atomic masses of all the different isotopes in a sample, with each isotope's contribution to the average determined by how big a fraction of the sample it makes up.
The relative atomic mass of Cadmium is 112.411 amu.

Which of the following is/are correct?

chemistry the language of chemistry atomic weight and molecular weight masses of atoms and molecules chemical analysis and formulae

  1. One atomic mass unit is a mass unit equal to exactly one -twelfth (1/12th) the mass of one atom of carbon-12.

  2. One atomic mass unit is a mass unit equal to exactly one - sixteenth (1/16th) the mass of one atom of oxygen-16.

  3. The relative atomic mass of the atom of an element is defined as the average mass of the atom, as compared to 1/12h the mass of one carbon-12 atom.

  4. None of the above.

Show answer
Correct Option: A,B,C
Explanation:

The following are correct. One atomic mass unit is a mass unit equal to exactly one -twelfth (1/12th) the mass of one atom of carbon-12.
One atomic mass unit is a mass unit equal to exactly one - sixteenth (1/16th) the mass of one atom of oxygen-16.
The relative atomic mass of the atom of an element is defined as the average mass of the atom, as compared to 1/12h the mass of one carbon-12 atom.
For example, the mass of one carbon-12 atom is 12 amu. The mass of one magnesium-24 atom is 24 amu. The mass of one calcium-40 atom is 140 amu.

Oxygen occurs in nature as a mixture of isotopes $^16O$, $^17O$ and $^18O$ having atomic masses of 15.995 u, 16.999 u and 17.999 u and relative abundance of 99.763%, 0.037%, and 0.200% respectively. What is the average atomic mass of oxygen?

chemistry the language of chemistry atomic weight and molecular weight masses of atoms and molecules chemical analysis and formulae

  1. 15.999 u

  2. 16.999 u

  3. 17.999 u

  4. 18.999 u

Show answer
Correct Option: A
Explanation:
 Average atomic mass of an element existing in different isotopes is given by:
$M _{ avg }=\dfrac { \sum _{ i=1 }^{ n }{ { M } _{ i }{ A } _{ i } }  }{ \sum _{ i=1 }^{ n }{ A _{ i } }  } $
where $M _i=$atomic mass of an isotope with relative abundance of $A _i$
Given:$M _1=15.995 u,A _1=99.763,M _2=16.999 u, A _2=0.037,M _3= 17.999 u, A _3=0.200$
on subtitutiing we get:
${ M } _{ avg }=\dfrac { 15.995\times 99.763+16.999\times 0.037+17.999\times 0.200 }{ 99.763+0.037+0.200 } $
${ M } _{ avg }=15.999\ u$
option A is correct

For every ,one $^{37}Cl$ isotope there are three $^{35}Cl$ isotopes, in a sample of chlorine. What will be the average atomic mass of chlorine?

chemistry the language of chemistry atomic weight and molecular weight masses of atoms and molecules chemical analysis and formulae

  1. 35

  2. 37

  3. 35.5

  4. 35.6

Show answer
Correct Option: C
Explanation:
Average atomic mass of an element existing in different isotopes is given by:
$M _{ avg }=\dfrac { \sum _{ i=1 }^{ n }{ { M } _{ i }{ A } _{ i } }  }{ \sum _{ i=1 }^{ n }{ A _{ i } }  } $
where $M _i=$atomic mass of an isotope with relative abundance of $A _i$
Given:$M _1=37 u,A _1=1,M _2=35 u, A _2=3$
on subtitutiing we get:
${ M } _{ avg }=\dfrac { 37\times 1+35\times 3 }{ 1+3 } $
${ M } _{ avg }=35.5\ u$
option C is correct

One gm metal $M^{3+}$ was discharged by the passage of $1.81\times 10^{23}$ electrons. What is the atomic mass of metal?

chemistry the language of chemistry atomic weight and molecular weight masses of atoms and molecules chemical analysis and formulae

  1. $8g/mol$

  2. $9g/mol$

  3. $10g/mol$

  4. None of these

Show answer
Correct Option: C
Explanation:
Applying faraday's first law-
$ \dfrac{Q}{F} = \dfrac{Wt}{Mwt}\times V.f.$
$ \dfrac{ne}{F}= \dfrac{1}{Mwt}\times 3$
$  Mwt         = \dfrac{1\times F\times 3}{ne}$
$  Mwt         = 10\dfrac{gm}{mol}$

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

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

  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

Show answer
Correct Option: A

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

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

  1. $5\times 10^{-5}$

  2. $2\times 10^{-5}$

  3. $3\times 10^{-5}$

  4. $5\times 10^{-6}$

Show answer
Correct Option: C
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.

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

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

  1. 30, 20, 45, 5

  2. 28, 1, 14, 57

  3. 24, 23, 12, 1

  4. None of above

Show answer
Correct Option: B
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 % $O = \dfrac{48.00 g}{84.01 g} \times 100 = 57.14$%

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

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

  1. $31\%$

  2. $43.58\%$

  3. $25\%$

  4. $55\%$

Show answer
Correct Option: C
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.