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Tricks Of Concept of Atomic, Molecular and Equivalent Masses for NDA

Atomic and Molecular Mass

Q 1713523449


NCERT Exemplar

Assertion : (A) One atomic mass unit is defined as one twelfth of the mass of one carbon-12 atom.

Reason : (R) Carbon-12 isotope is the most abundant isotope of carbon and has been chosen as standard

(A) Both A and R individually true and R is the correct explanation of A
(B) Both A and R are individually true but R is not the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Solution:

Both assertion and reason are true but reason is not the correct explanation of assertion

atomic masses of the elements obtained by scientists by comparing with the mass of carbon comes out to be close to whole number value
Correct Answer is `=>` (B)
Q 2807001888

The statement which is wrong about gram atomic mass is

(A)

it is the atomic mass expressed in gram

(B)

it is also called gram atom

(C)

one gram atom of an element contain `6xx10^(23) ` atoms

(D)

None of the above

Solution:

1 g-atom = N atoms =` 6.023xx10^(23)` atoms = g - atomic weight

e.g. 1 g atom of oxygen = N atoms of oxygen

` = 6.023xx10^(23)` atoms of oxygen = 16 g
Correct Answer is `=>` (D) None of the above
Q 2827101981

Chlorine occurs in nature in the form of two isotopes with atomic mass `35` and `37` in the ratio of `3 : 1` respeetively. The average
atomie mass of chlorine is

(A)

`38.5`

(B)

`35.5`

(C)

`36`

(D)

None of these

Solution:

`M_(av) = ( 35xx3+37xx1)/(3+1) = 35.5`
Correct Answer is `=>` (B) `35.5`
Q 2817356289

Statement I : Average atomic mass of elements may be in fraction.

Statement II. Due to presence of isotopes of elements.

(A)

Both the statements are true and Statement II is the correct explanation of Statement I

(B)

Both the statements are true but Statement II is not the correct explanation of Statement I.

(C)

Statement I is true, but Statement II is false

(D)

Statement I is false, but Statement II is true.

Solution:

`M_(av) = ( 35xx3+37xx1)/(3+1) = 35.5`
Correct Answer is `=>` (A) Both the statements are true and Statement II is the correct explanation of Statement I
Q 2837801782

`1 u` is equal to

(A)

`1/12 ` of `C^(12)``

(B)

`1/14` of `O^(16)`

(C)

`1 g` of `H_2`

(D)

`1.66xx10^(-23) kg`

Solution:

`M_(av) = ( 35xx3+37xx1)/(3+1) = 35.5`
Correct Answer is `=>` (B) `1/14` of `O^(16)`

Equivalent Mass and Vapour Density

Q 2817801780

Number of atoms present in a molecule is called

(A)

mole ratio

(B)

molecularity

(C)

atomicity

(D)

Avogadro's number

Solution:

`M_(av) = ( 35xx3+37xx1)/(3+1) = 35.5`
Correct Answer is `=>` (C) atomicity
Q 2837101982

If M is the molecular mass of `KMnO_4` then equivalent weight of `KMnO_4` in acidic medium is

(A)

`M/2`

(B)

`M/4`

(C)

`M/7`

(D)

`M/5`

Solution:

Eq. wt. of an oxidising agent ` = (text{Molar mass}(M))/text(Change in oxidation number) = M/5`
Correct Answer is `=>` (D) `M/5`
Q 2847101983

What is the equivalent mass of `KMnO_4` when it change into
`Mn_2(SO_4)_3?`

(A)

`M`

(B)

`M/5`

(C)

`M/6`

(D)

`M/4`

Solution:

Eq. wt. of an oxidising agent ` = (text{Molar mass}(M))/text(Change in oxidation number) = M/5`
Correct Answer is `=>` (D) `M/4`
Q 2807101988

Approximate atomie weight of a metal is `26.89`. If its equivalent weight is `8.9` its exact atomic weight will be

(A)

`26.7`

(B)

`8.9`

(C)

`26.89`

(D)

`17.8`

Solution:

Eq. wt. of an oxidising agent ` = (text{Molar mass}(M))/text(Change in oxidation number) = M/5`
Correct Answer is `=>` (A) `26.7`
Q 2827112081

Potassium permanganate gives the following reactions in neutral `mnO_4^- + 2H_2O +3e^- → MnO_2+4OH^-` The equivalent weight of `KMnO_4` is (atomic mass of `Mn = 55u`)

(A)

`158`

(B)

`79`

(C)

`52.66`

(D)

`31.6`

Solution:

`overset(+7)MnO_4^- +2H_2O+3e^- → overset(+4)MnO_2+4OH^-`

Change in oxidation number = 3

Equivalent weight of `KMnO_4 = 158/3 = 52.66`
Correct Answer is `=>` (C) `52.66`
Q 2847112083

Equivalent weight of crystalline oxalic acid is

(A)

`45`

(B)

`90`

(C)

`126`

(D)

`63`

Solution:

Eq . wt . of an acid = `text(mol . wt)/text(Basicity of acid)`

` = 126/2 = 63`
Correct Answer is `=>` (D) `63`
Q 2867112085

Equivalent weight of nitrogen varies in its oxides, because it

(A)

contains five electrons in its valence orbit

(B)

contains half filled p -orbitals

(C)

is a diatomic molecule

(D)

has variable valency

Solution:

Eq . wt . of an acid = `text(mol . wt)/text(Basicity of acid)`

` = 126/2 = 63`
Correct Answer is `=>` (D) has variable valency
Q 2857212184

Atomic weight of a trivalent element of equivalent weight 9 is

(A)

`9`

(B)

`27`

(C)

`18`

(D)

`36`

Solution:

At. wt. of an element = eq. wt `xx` valency = `9 xx 3 = 27`
Correct Answer is `=>` (B) `27`
Q 2887212187

A reaction between `HCI` and `O_2` is given by `4HCl +O_2 → 2H_2O+2Cl_2` The equivalent weight of `HCl` is equal to

(A)

its molecular weight

(B)

half of its molecular weight

(C)

twice of its molecular weight

(D)

four times its molecular weight

Solution:

At. wt. of an element = eq. wt `xx` valency = `9 xx 3 = 27`
Correct Answer is `=>` (A) its molecular weight
Q 2817312280

Equivalent weight of sulphur in `SCl_2` is `16`. What is the equivalent weight of `S` in `S_2Cl_2`

(A)

`16`

(B)

`64`

(C)

`32`

(D)

`8`

Solution:

At. wt. of an element = eq. wt `xx` valency = `9 xx 3 = 27`
Correct Answer is `=>` (C) `32`
Q 2807312288

Equivalent Weight of a metal is `29.4`. It forms metal sulphate isomorphous with epsom salt. The atomic weight of the metal is

(A)

`58.8`

(B)

`14.7`

(C)

`29.4`

(D)

`88.2`

Solution:

Atomic wt. of a metal = eq. wt `xx` valency= `29.4 xx 2 = 58.8`
Correct Answer is `=>` (A) `58.8`
Q 2867356285

Which of the following statements are true Select the correct answer using the codes given below.
I. The valencies of elements forming Isomorphous compounds are same.
II. Equivalent mass may vary with change of valency.
III. Some elements show variable valency.


(A)

I and II

(B)

II and Ill

(C)

I and Ill

(D)

All of these

Solution:

Atomic wt. of a metal = eq. wt `xx` valency= `29.4 xx 2 = 58.8`
Correct Answer is `=>` (D) All of these
Q 2827456381

Statement I : Equivalent mass of element may vary.
Statement.II : Valency of element may vary.

(A)

Both the statements are true and Statement II is the correct explanation of Statement I

(B)

Both the statements are true but Statement II is not the correct explanation of Statement I.

(C)

Statement I is true, but Statement II is false

(D)

Statement I is false, but Statement II is true.

Solution:

Atomic wt. of a metal = eq. wt `xx` valency= `29.4 xx 2 = 58.8`
Correct Answer is `=>` (A) Both the statements are true and Statement II is the correct explanation of Statement I
Q 2877101986

`74 g` of a metallic ehloride contains `35.5 g` of chlorine. The equivalent weight of tho metal is

(A)

`38.5`

(B)

`74.4`

(C)

`35.5`

(D)

`71`

Solution:

The number of parts of a substance that combines with `35.5` parts by mass of chlorine is called the equivalent mass
of the substance. Therefore, equivalent weight (mass) of the metal is `74g - 35.5 g = 38.5 g`
Correct Answer is `=>` (A) `38.5`

Mole Concept

Q 2817312289

The mass of an atom of nitrogen is

(A)

`14/(6.022 xx 10^(22)) g`

(B)

`28/(6.022xx 10^(23)) g`

(C)

`1/(6.022xx10^(23)) g`

(D)

`14 u`

Solution:

Mass of an atom of an element `= text(Molar mass of an atom of element)/(6.023xx10^(23))`

` = 14/(6.022xx10^(23)) g`
Correct Answer is `=>` (A) `14/(6.022 xx 10^(22)) g`
Q 1783512447

16 g of oxygen has same number of molecules as in
NCERT Exemplar
(A)

16g of CO

(B)

28 g of `N_2`

(C)

14 g of `N_2`

(D)

1.0g of `H_2`

Solution:

The number of molecules can be calculated as follows

Number of molecules = `text(mass)/text(molar mass)xxtext(avogadro number)`


Number of molecules, in 16 g oxygen`= 16/32xxN_A = N_A/2`

In 16 g of CO `= 16/28xxN_A = N_A/1.75`

In 28 g of `N_2= 28/28xxN_A = N_A`

In 14 g of `N_2 = 14/28xxN_A = N_A/2`


In 1 g of `H_2 = 1/2xxN_A = N_A/2`

So, 16 g of `O_2 = 14`g `text(of ) N_2 = 1.0g text(of) H_2`
Correct Answer is `=>` (C)
Q 2847412383

How many atoms are present in a mole of `H_2SO_4 ?`

(A)

`3xx6.02xx10^(23)`

(B)

`5xx6.022xx10^(23) `

(C)

`6xx6.02xx10^(23)`

(D)

`7xx6.02xx10^(23)`

Solution:

1 mole `H_2SO_4 ` = 2 mole of H atoms +1 mole of S atom +4 mole of O atoms .

` = 7` mole atoms

` = 7xx6.023xx10^(23)` atoms
Correct Answer is `=>` (D) `7xx6.02xx10^(23)`
Q 1773101946

One mole of oxygen gas at STP is equal to ........ .
NCERT Exemplar

(This question may have multiple correct answers)

(A) `6.022xx10^(23)` molecules of oxygen
(B) `6.022xx10^(23)` atoms of oxygen
(C) `16 g` of oxygen
(D) `32 g` of oxygen
Solution:

1 mole of `O_2` gas at STP `= 6.022xx10^(23)` molecules of `O_2`(avogadro number ) = 32 g of `O_2`
Hence 1 mole of oxygen gas is equal to molecular weight of oxygen as well as avogadro number
Correct Answer is `=>` (A)
Q 2887412387

`2 g` of oxygen contain number of atoms equal to that contained in

(A)

`0.5 g` hydrogen

(B)

`4.0 g` sulphur

(C)

`7.0 g` nitrogen

(D)

`2.3 g` sodium

Solution:

Equal number of moles contain equal number of atoms

`2g` of oxygen ` = 2/16` mole ` = 0.125` mole

Similarly `4 g` sulphur `= 4/32 = 2/16` mole `= 0.125` mole

`0.5 g` hydrogen `= 0.5/1 = 0.5` mole

`7.0 g` nitrogen `= 7/14 = 0.5` mole

`2.3 g` sodium `= 2.3/23 = 0.1` mole

Therefore atoms in `2 g` oxygen = atoms in `4.0 g` sulphur.
Correct Answer is `=>` (B) `4.0 g` sulphur
Q 1703312248

Which of the following pairs have the same number of atoms?
NCERT Exemplar
(A)

`16` g of `O_2`(g) and 4g of `H_2`

(B)

`16`g of `O_2` and `44` g of `CO_2`

(C)

`28` g of `N_2` and `32` g of `O_2`

(D)

`12` g of C(s) and `23` g of Na(s)

Solution:

Number of atoms in 28 g of `N_2` = `(28)/(28)xxN_Axx2 = 2N_A` (where `N_A` = avogadro number)



Number of atoms in 32g of `O_2` = `(32)/(32)xxN_Axx2 = 2N_A`



(d) 12 g of C(s) contains atoms `= 12/12xxN_Axx1 = N_A`


Number of .atoms in 23 g of Na(s) `= 23/23xxN_Axx1 = N_A`

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