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

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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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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

(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`

Correct Answer is `=>` ()

Q 2827512481

The number of sulphur atoms in its `40 g` is

(A)

`40xx6.022xx10^(23)`

(B)

`32xx6.022xx10^(23)`

(C)

`(40xx6xx10^(23))/32`

(D)

`(32xx6xx10^(23))/40`

Solution:

`40 g` sulphur ` = 40/32` mole

`1` mole `= 6.023 xx 10^(23)` atoms

`40/32 ` mole ` = ( 40xx6.023xx10^(23))/32` atoms

Correct Answer is `=>` (C) `(40xx6xx10^(23))/32`

Q 2817145989

A mole of compound is composed of `6.023xx10^(23)` atoms of hydrogen,`35.5 g` of chlorine and `48 g` of oxygen. The compound is

(A)

`HClO`

(B)

`HClO_2`

(C)

`HClO_3`

(D)

`HClO_4`

Solution:

`100` molecules of `H_2`

`6.023xx10^(23)` atom of `H = 1` mole `H` atom

`35.5 g` chlorine = `1` mole `Cl` atom
`48 g` oxygen = `3` mole `O` atoms
Chemical formula of the compound
`=HCIO_3`

Correct Answer is `=>` (C) `HClO_3`

Q 1753201144

If the concentration of glucose `(C_6H_12O_6 )` in blood is `0.9 g L^(-1)`

what will be the molarity of glucose in blood?
NCERT Exemplar

(A)

`5 M`

(B)

`50 M`

(C)

`0.005 M`

(D)

`0.5 M`

Solution:

In the given question. `0.9 g L^(-1)` , means that `1000 ml (or 1 L)` solution contains 0.9 g of
glucose.

Number of moles = 0. 9 g glucose `= (0.9)/180` mol glucose

` = 5 xx 10^(- 3)` mol glucose

(where, molecular mass of glucose `(C_6H_10_6 ) = 12 xx 6 + 12 xx 1 + 6 xx 16 = 180 u)`
i.e., 1 L solution contains 0.05 mole glucose or the molarity of glucose is 0.005 M.

Correct Answer is `=>` (C) `0.005 M`

Q 2817156089

How many moles of electrons weigh one kilogram?

(A)

`6.023xx10^(23)`

(B)

`1/(9.08) xx 10^(31)`

(C)

`(6.023)/(9.108) xx 10^(54)`

(D)

`1/(9.108xx6.023) xx 10^(8)`

Solution:

`9.108xx10^(-31) kg` = mass of 1 electron

`1 kg = 1/(9.108xx10^(-31))`

` = ( 1xx10^(31))/(9.108)` electrons

`6.023xx10^(23)` electrons = 1 mole ` (1xx10^(31))/(9.108)`

` = (1xx10^(31))/(9.108xx6.023xx10^(23))` moles

` = 10^8/(9.108xx6.023) ` moles of electrons

Correct Answer is `=>` (D) `1/(9.108xx6.023) xx 10^(8)`

Q 2887356287

Consider the following statements. Select the correct answer using the codes given below Codes
I. The number of moles is directly proportional t:o the number of molecules of the substance.
II. Mole concept is also applicable to ionic compounds, which do not contain molecules.
III. Molecular mass' word is not suitable word for the ionic compounds.
IV. Formula mass unit is taken for ionic compounds

(A)

I, II and Ill

(B)

II, Ill and IV

(C)

I, Ill and IV

(D)

All of these

Solution:

Correct Answer is `=>` (D) All of these

Q 2017667580

Match the following

Column I		Column II
(A)	88 g of `CO_2`	(1)	0.2 mol
(B)	`6.022 xx 10^23` molecules of `H_2O`	(2)	2 mol
(C)	5.6 L of `O_2` at STP	(3)	1 mol
(D)	96 g of `O_2`	(4)	`6.022 xx 10^23` molecules
(E)	1 mole of any gas	(5)	3 mol

NCERT Exemplar

Solution:

`A.->(2) B -> (3) C.-> (1) D.-> (5) E.-> (4)`

A. Number of moles of `CO_2` molecule = `(text(weight in gram of ) CO_2)/(text(molecular weight of) CO_2) = 88/44 = 2` mol

B. 1 mole of a substance `= N_A` molecules = `6.022xx10^(23)` molecules
= avogadro Number

= `6.022xx10^(23)` molecules of `H_2O` = 1 mol

C. 22.4L of `O_2` at STP = 1 mol

5.6L of `O_2` at STP `5.6/22.4 mol`

D. Number of moles of 96 g of `O_2` = `96/32 = 3` mol

E. 1 mole of any gas = avogadro Number = `6.022xx10^(23)` mplecules

Q 2837512482

Number of atoms in `4.25 g` of `NH_3` is (approx.)

(A)

`1 xx 10^(23)`

(B)

`1.5 xx 10^(23)`

(C)

`2 xx 10^(23)`

(D)

`6xx10^(23)`

Solution:

`4.25 g NH_3 = (4.25)/17 ` = mole ` NH_3`

` (4xx4.25 xx 6.023 xx 10^(23))/17 = 6.023xx10^(23)` atoms
(1 mole `NH_3` =, 1 mole N atoms + 3 mole H atoms = 4 mole atoms)

Correct Answer is `=>` (D) `6xx10^(23)`

Q 2857512484

Which of the following has maximum number of atoms?

(A)

`18 g`of water

(B)

`16 g` of `O_2`

(C)

`4.4 g` of `O_2`

(D)

`16 g` of `CH_4`

Solution:

`18 g` water `= 18/18 = 1` mole

` = 1xx3xx6.023xx 10^(23)` atoms

`16 g O_2 = 16/32 = 0.5` mole

` = 0.5xx2xx6.023xx10^(23)` atoms

`4.4 g O_2 = (4.4)/32 = 0.1375` mole

` = 0.1375 xx 2 xx 6.023xx10^(23)` atoms

`16g CH_4 = 16/16 = 1` mole

` = 1xx5xx6.023xx10^(23)` atoms

Correct Answer is `=>` (D) `16 g` of `CH_4`

Q 1904723658

Which one of the following will have largest
number of atoms?

(i) 1 g of Au (s)
(ii) 1 g of Na (s)
(iii) 1 g of Li (s)
(iv) 1 g of `Cl_2` (g)
Class 11 Exercise 1 Q.No. 28

Solution:

`1` g Au `= 1/197` mol `=1/197 xx 6.02 xx 10^23` atoms
.
(ii) `1` g Na `=1/23` mol `=1/23 xx 6.02 xx 10^23` atoms

`= 0.261 xx 10^23` atoms `= 2.61 xx 10^22` atoms

(iii) `1` g Li `1/7` mol = `1/7` mol `= 7 xx 6.02 xx 10^22` atoms

`=0.86 xx 10^23` atoms `=0.86 xx 10^22 ` atoms

(iv) `1` g `Cl_2`

`1/7` mol `=1/7 xx 6 .02 xx 10^23` molecules

`= 71 xx 6.02 xx 10^23` atoms

`= 0.69 xx 10^22` atoms

Thus 1 g of Li has the largest number of
atoms.

Q 2817612580

The mass of a molecule of water is

(A)

`3xx10^(-26) kg`

(B)

`3xx10^(-25) kg`

(C)

`1.5 xx 10^(-25) kg`

(D)

`2.5xx10^(-26) kg`

Solution:

Mass of molecule of `H_2O`

` = 18/(6.023xx10^(23)) g`

` = 29885xx2`

` = 29885xx10^(-26) kg`

`3xx10^(-26) kg`

Correct Answer is `=>` (A) `3xx10^(-26) kg`

Q 1713401340

One mole of any substance contains `6.022 xx 10^(23)` atoms/molecules.
Number of molecules of `H_2SO_4` present in 100ml of 0.02M `H_2SO_4` solution is ...................
NCERT Exemplar

(A)

`12.044xx10^(20)` molecules

(B)

`6.022xx10^(23)` molecules

(C)

`1xx10^(23)` molecules

(D)

`12.044xx10^(23)` molecules

Solution:

One mole of any substance contains `6.022xx10^(23)` atoms /molecules

Hence, Number of millimoles of `H_2SO_4`

= molality `xx`volume in mL

`= 0.02xx100 = 2` millimoles

`= 2xx10^(-3)` mol

Number of molecules= number of `text(moles) xx N_A`

`= 2xx10^(-3)xx6.022xx10^(23)`

` = 12.044xx10^(20)` molecules

Correct Answer is `=>` (A) `12.044xx10^(20)` molecules

Q 2817045889

Which of the following has largest number of atoms?

(A)

`71 g` of chlorine

(B)

`48 g` of magnesium

(C)

`127 g` of iodine

(D)

`4 g` of hydrogen

Solution:

`71g` chlorine `= 71/71 = 1` mole `Cl_2`

` = 2 xx6.023xx10^(23) ` atoms

`48g` magnesium `= 48/24 = 2` mole `Mg`
` = 2xx6.023xx10^(23) ` atoms

`127 g` iodine `= 127/(127 xx 2) = 0.5` mole `I_2`
` = 2xx0.5xx6.023xx10^(23)` atoms

`4g` hydrogen `= 4/2 =2 ` mole `H_2`

` = 2xx2xx6.023xx10^(23) ` atoms (maximum atoms)

Correct Answer is `=>` (D) `4 g` of hydrogen

Q 1703201148

What will be the molality of the solution containing 18.25 g of HCl gas in
500 g of water?
NCERT Exemplar

(A)

0.1 m

(B)

1 M

(C)

0.5 m

(D)

1 m

Solution:

Molality is defined as the number of moles of solute present in 1 kg of solvent. It is
denoted by m.

`text(Molality(m))` = `text(Moles of solute)/text(Mass of solvent (in kg)` ...................(i)

Given that, Mass of solvent `(H_2O) = 500g = 0.5kg`

Weight of HCI `= 18.25 g`

Molecular weight of HCI `= 1 xx 1 + 1 xx 35.5 = 36.5g`

Moles of HCI `= (18.25)/(36.5) = 0.5`

`m = 0.5/0.5 = 1m`

Correct Answer is `=>` (D) 1 m

Q 2857145984

The number of moles of `CO_2` whieh eontain `16 g` of oxygen is

(A)

`0.25`

(B)

`0.50`

(C)

`1`

(D)

`2`

Solution:

`32 g` oxygen present in `1` mole `CO_2` therefore `16g` oxygen present in = `(16xx1)/(32) = 0.5` mole `CO_2`

Correct Answer is `=>` (B) `0.50`

Q 2887256187

Number of electron present in `10 g` of `H_2O` is

(A)

`(6.022xx10^(-23))/6`

(B)

`(6.02xx10^(25))/18`

(C)

`(6.022xx10^(22))/6`

(D)

None of these

Solution:

`18g H_2O` contains

` = 10` mole electrons

` 10xx6.023xx10^(23)` electrons

`10 g H_2O` will contain

` = (10xx10xx6.023xx10^(23))/18` electrons

` = (6.023xx10^(25))/18` electrons

Correct Answer is `=>` (B) `(6.02xx10^(25))/18`

Q 2817256189

The volume occupied by `4.4 g` of `CO_2` at STP is

(A)

`22.4 L`

(B)

`2.24 L`

(C)

`0.224 L`

(D)

`0.1 L`

Solution:

Volume occupied by `1` mole `(44g)`
`CO_2 = 224 L` ( at STP)

Hence `(4.4)/(44) ` mole `CO_2` will occupy

` = (224xx4.4)/(44) = 2.24 L`

Correct Answer is `=>` (B) `2.24 L`

Q 1783101047

The number of atoms present in one mole of an element is equal to Avogadro number. Which of the following element contains the greatest number of atoms?

NCERT Exemplar

(A)

`4 g` He

(B)

`46 g` Na

(C)

`0.40 g` Ca

(D)

`12 g` He

Solution:

For comparing number of atoms, first we calculate the moles as all are monoatomic
and hence, `text(moles) xx N_A` = number of atoms

Moles of 4 g He `= 4/4 = 1` mol

46 g Na = `46/23 = 2 mol`

0.40 g Ca = `0.40/40 = 0.1 mol`

12 g He = `12/4 = 3 mol`

Hence, 12 g He contains greatest number of atoms as it possesses maximum number
of moles.

Correct Answer is `=>` (D) `12 g` He

Chemical Equation

Q 2837356282

Consider the following equation for the formation of ammonia from nitrogen and hydrogen `N_2+3H_2 → 2NH_3` How many hydrogen molecules are required to react with `100` molecules of nitrogen?

(A)

`100`

(B)

`200`

(C)

`300`

(D)

`400`

Solution:

`underset(1)(N_2)+ underset(3)(3H_2) → 2NH_3`

`1` molecules of `N_2` reacts with 3 molecules of `H_2`

`100` molecules of `N_2 = 300` molecules of `H_2`

Correct Answer is `=>` (C) `300`

Q 1723623541

NCERT Exemplar

Assertion : (A) Combustion of 16 g of methane gives 18 g of water

Reason : (R) In the combustion of methane, water is one of the products.

(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:

Assertion is false but Reason is true.
Combustion of 16 g of methane gives 36 g of water.

`CH_4+2O_2rightarrowCO_2+2H_2O`

Correct Answer is `=>` (D)

Q 1783112047

Sulphuric acid reacts with sodium hydroxide as follows

`H_2SO_4+2NaOH rightarrow Na_2SO_4+2H_2O`

When 1L of 0.1M sulphuric acid solution is aLLowed to react with 1L of
0.1M sodium hydroxide solution, the amount of sodium sulphate formed
and its molarity in the solution obtained is
NCERT Exemplar

(This question may have multiple correct answers)

(A) `0.1mol L^(-1)`

(B) `7.10g`

(D) `3.55g`

Solution:

For the reaction

`H_2SO_4+2NaOH rightarrowNa_2SO_4+2H_2O`

1 L of 0.1 M `H_2SO_4` contains= 0.1 mole of `H_2SO_4`

1 L of 0.1 M `NaOH` contains= 0.1 mole of `NaOH`

According to the reaction, 1 mole of `H_2SO_4` reacts with 2 moles of `NaOH`. Hence, 0.1 mole of
`NaOH` will react with 0.05 mole of `H_2SO_4` (and 0.05 mole of `H_2SO_4` will be left unreacted), i.e.,
`NaOH` is the limiting reactant. Since, 2 moles of `NaOH` produce 1 mole of `NaOH` .
Hence, 0.1 mole of `NaOH` will produce 0.05 mole of `Na_2SO_4`

Mass of `Na_2SO_4` = moles `xx` molar mass

`= 0.5xx(46+32+64)g`

`= 7.10g`

Volume of solution after mixing = 2 L

Since, only 0.05 mole of `H_2SO_4` is left behind as `NaOH` completely used in the reaction.
Therefore, molarity of the given solution is calculated from moles of `H_2SO_4` .

`H_2SO_4` lett unreacted in the solution = 0.05 mole
Molarity of the solution= `00.5/2` `= 0.025molL^(-1)`

Correct Answer is `=>` (B)

Q 2857356284

What weight of `SO_2` can be made by burning sulphur in `5.0` moles of oxygen ?

(A)

`640 g`

(B)

`160 g`

(C)

`80 g`

(D)

`320 g`

Solution:

`S+ underset(1 mol e)(O_2) → underset(32+2xx16 = 64 g) (SO_2)`

`1` mole `O_2` gives `= 64 g SO_2`

Therefore `5` mole `O_2` will give ` = 64xx5 = 320 g SO_2`

Correct Answer is `=>` (D) `320 g`

Q 1783701647

Which of the following statements is correct about the reaction given
below?

`4Fe(s) + 3O_2(g)rightarrow 2Fe_2O_3(g)`

NCERT Exemplar

(A)

Total mass of iron and oxygen in reactants= total mass of iron and oxygen in product therefore it follows law of conservation of mass

(B)

Total mass of reactants = total mass of product, therefore, law of multiple proportions is followed

(C)

Amount of `Fe_2O_ 3` can be increased by taking any one of the reactants (iron or oxygen) in excess

(D)

Amount of `Fe_2O_3` produced will decrease if the amount of any one of the reactants {iron or oxygen) is taken in excess

Solution:

According to the law of conservation of mass,
Total mass of reactants = Total mass of pmducts
Amount `Fe_2O_3`of is decided by limiting reagent

Correct Answer is `=>` (A) Total mass of iron and oxygen in reactants= total mass of iron and oxygen in product therefore it follows law of conservation of mass

Strength of a Solution

Q 1723512441

Which of the following solutions have the same concentration ?
NCERT Exemplar

(This question may have multiple correct answers)

(A) 20 g of NaOH in 200 mL of solution

(B) 0.5 mol of KCI in 200 mL of solution

(D) 20 g of KOH in 200 mL of solution

Solution:

(a) Molality(M) = `(text(weight of) NaOHxx1000)/(text(Molecular weight of) NaOH xxV(mL))`

`= (20xx1000)/(40xx200) = 2.5 M`

(b) `M = (0.5xx1000)/(200) = 2.5M`

(c) `M = (40xx1000)/(10xx100) = 10M`

(d) `M = (20xx1000)/(56xx200) = 1.785M`

Thus, 20 g NaOH in 200 ml of solution an;'J 0.5 mol of KCI in 200 ml have the same
concentration

Correct Answer is `=>` (A)

Q 1712191939

If 500 ml of a 5M solution is diluted to 1500 ml, what will be the
molarity of the solution obtained ?

NCERT Exemplar

(A)

`1.5 M`

(B)

`1.66 M`

(C)

`0.017 M`

(D)

`1.59 M`

Solution:

Given that `M_1 = 5M`

`V_1 = 500mL`

`V_2 = 1500mL`

`M_2 = M`
For dilution, a general formula is

`M_1V_1 =M_2V_2`

`500xx5M = 1500xxM`

`M = 5/3 = 1.66M`

Correct Answer is `=>` (B) `1.66 M`

Q 1702091838

What will be the molarity of a solution, which contains 5.85 g of NaCl(s)
per 500 ml?
NCERT Exemplar

(A)

`4 molL^(-1)`

(B)

`20 molL^(-1)`

(C)

`0.2 molL^(-1)`

(D)

`2 molL^(-1)`

Solution:

Since, molarity (M) is calculated by following equation

Molarity = `(text(weight)xx100)/(text(molecular weight) xx volume (mL))`

= `(5.85xx1000)/(58.5xx500) = 0.2 molL^(-1)`

`text(Note)` Molarity of solution depends upon temperature because volume of a solution is
temperature dependent.

Correct Answer is `=>` (C) `0.2 molL^(-1)`

Q 2827556481

Match List I with List II and select the correct answer from the codes given below the lists

List I		List II
(A)	molarity	(1)	Gram formula weight of solute in one litre of solution.
(B)	Molality	(2)	Number of gram equivalents of solute dissolved in one litre of solution
(C)	Normality	(3)	Number of moles of solute dissolved in 1 kg of solvent .
(D)	Formality	(4)	Number of moles of solute in one litre of solution

(A)

`A →4 , B → 3, C → 2 , D → 1`

(B)

`A → 1 , B → 2 , C → 3 , D → 4`

(C)

`A → 4 , B → 2 , C → 3 , D → 1`

(D)

`A → 2 , B → 3 , C → 4 , D → 1`

Solution:

Correct Answer is `=>` (A) `A →4 , B → 3, C → 2 , D → 1`