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Atomic weight and molecular weight - class-XII

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Boron has two stable isotopes, $^{10}B (19\%)$ and  $^{11}B(81\%)$. Average atomic weight for boron in the periodic table is:
  1. 10.8

  2. 10.2

  3. 11.2

  4. 10.0


Correct Option: A
Explanation:
Average atomic weight $=\dfrac{ \sum \%abundant \times atomic\ mass}{100}$
$=\dfrac{ 19 \times 10 + 81 \times 11}{100}=10.81$

Option B is the answer.

An element $X$ have three isotopes $X^{20}, X^{21}$ and $X^{22}$. The percentage abundance of $X^{20}$ is $90\%$ and its average atomic mass of elements is $20.18$. The percentage abundance of $X^{21}$ should be ______________.

  1. $2\%$

  2. $8\%$

  3. $10\%$

  4. $0\%$


Correct Option: A
Explanation:

Average atomic mass $=\dfrac {\displaystyle \sum \%\ abundance\ \times \ A.M }{100}$

$20.18=\dfrac {20\times 90+x\times21+(10-x)22}{100}$

$2018=1800+21x+220-22x$

$x=2020-2018$

$x=2\%$

Option $(A).$

The fractional abundance of $Cl^{35}$ in a sample of chlorine containing only $Cl^{35}$ (atomic weight $=34.9$) and $Cl^{37}$ (atomic weight $=36.9$) isotopes, is $0.6$. The average mass number of chlorine is  _______________.

  1. $35.7$

  2. $35.8$

  3. $18.8$

  4. $35.77$


Correct Option: A
Explanation:

$\dfrac {C1^{35}}{C1^{37}}\Rightarrow \dfrac {0.6}{(1-0.6)}=\dfrac {0.6}{0.4}$

Average atomic mass $=\dfrac {34.9\times 0.6+36.9\times 0.4}{1}$

Average Atomic mass $=35.7$

Option $A$ is the answer.

The $O^{18}/O^{16}$ ratio in some meteorites is greater than the used to calculate the average atomic mass of oxygen one earth. The average mass of an atom of oxygen in these meteorites is ______ that of terrestrial oxygen atom?

  1. equal to

  2. greater than

  3. less than

  4. none of these


Correct Option: A

In the atomic weight determination. Dalton suggest the formula of water as $HO$ and the composition of water as hydrogen $=12.5\%$ and oxygen $=87.5\%$ by weight. What should be the atomic weight of oxygen on $H-$scale, on the basis of this information?

  1. $16$

  2. $8$

  3. $14$

  4. $7$


Correct Option: A

The isotopes of chlorine with mass numbers $35$ and $37$ whose average mass is 35.5 exist in the ratio of:

  1. $1:1$

  2. $3:1$

  3. $3:4$

  4. $3:2$


Correct Option: B
Explanation:

Let the fraction of $Cl^{35}$ isomer be $p$. The fraction of $Cl^{37}$ isomer will be $1-p$.

Average atomic mass $=\displaystyle {p\times m _{1}+(1-p)m _{2}}$
$\therefore 35.5 = p \times 35 + (1 - p) \times 37$

So, $p = 0.75$

Hence, the ratio of two isotopes will be $p: (1-p) = 0.75 : (1-0.75) = 0.75:0.25 = 3:1$

Option B is correct.

The relative atomic mass of naturally occurring chlorine is not a whole number. What is the reason for this ?

  1. Chlorine atoms can have different number of neutrons

  2. Naturally occurring chlorine cannot be obtained pure

  3. Chlorine is unstable

  4. The mass of the electrons has been included


Correct Option: A
Explanation:

 The relative atomic mass of an element is the average of the atomic masses of all the chemical element's isotopes as found in a particular environment, weighted by isotopic abundance. 


Chlorine has two isotopes available in Nature. 

One has At.mass = 35 which forms roughly $75\%$ of the naturally available Chlorine. 

Another has At.mass = 37 which forms roughly $25\%$ of the naturally available Chlorine. 

So they have fractional atomic mass.
Hence, the correct option is $\text{A}$

Among the following quantities:
(i) mass number
(ii) average mass of a Carbon atom in amu
(iii) the charge of nucleus in amu and
(iv) mass of a Carbon -12 atom in grams

  1. only (i) is whole number

  2. only (i) and (iii) are whole number

  3. only (i) and (iv) are whole numbers

  4. only (iii) and (iv) are whole numbers


Correct Option: B
Explanation:

(1) The mass number is always the whole number because the mass number is the total number of neutrons and protons in the nucleus of the atom and an atom never contains fractional or an integer number of protons or neutrons.

(2) Average mass of carbon atom in $amu$ is $12.011u$. It is in decimal and can be written as fractional form.
(3) Charge on the nucleus in $amu$:- Charge on the nucleus is equal to the number of protons and the number of protons are always whole.
(4) Mass of carbon atom= $\cfrac {Mass\quad of\quad carbon}{Avogadro\quad number}$
$=\cfrac {12g}{6.022\times 10^{23}}$
$=1.99\times 10^{-23}$ grams
It is not a whole number.

What is the approximate molecular mass of dry air containing 78% N$ _2$ and 22% O$ _2$?

  1. 48.88

  2. 18.88

  3. 28.88

  4. 38.88


Correct Option: C
Explanation:

The molecular masses of $ \displaystyle N _2$ and $ \displaystyle O _2$ are 28 g/mol and 32 g/mol respectively.
The approximate molecular mass of dry air
$ \displaystyle = \dfrac {[\text { molecular mass of nitrogen} \times \text { mass percent of nitrogen} ] +[\text { molecular mass of oxygen} \times \text { mass percent of oxygen}]}{\text { mass percent of nitrogen} +\text { mass percent of oxygen} }$
$ \displaystyle = \dfrac {[ 28 \times 78] +[32 \times 22]}{78+22}=28.88$ g/mol

What weight of sodium contains the same number of atoms as those in $8$ grams of oxygen?

  1. $10.5 g$

  2. $13.5 g$

  3. $11.5 g$

  4. $14.2 g$


Correct Option: C
Explanation:
Atomic weights of sodium and oxygen are 23 g/mol and 16 g/mol respectively.

8 g oxygen $ \displaystyle = \dfrac {8 \ g}{16 \ g/mol}= 0.5 \ $ moles of oxygen atoms.

Number of moles of sodium $ \displaystyle =$ number of moles of oxygen atoms $ \displaystyle = $ 0.5.

Weight of sodium $ \displaystyle = 0.5 \ mol \times 23 \ g/mol = 11.5 \ g$.

Note: Since sodium and oxygen samples have same number of atoms, they will have same number of moles.

What is the atomic mass of sodium?

  1. $22\ g/mol$

  2. $23\ g/mol$

  3. $11\ g/mol$

  4. $20\ g/mol$


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?

  1. 112.411 amu

  2. 20.00 amu

  3. 21.00 amu

  4. 20.43 amu


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?

  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.


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?

  1. 15.999 u

  2. 16.999 u

  3. 17.999 u

  4. 18.999 u


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?

  1. 35

  2. 37

  3. 35.5

  4. 35.6


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?

  1. $8g/mol$

  2. $9g/mol$

  3. $10g/mol$

  4. None of these


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}$
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