Some Basic Concepts of Chemistry

1!
Suppose the elements X and Y combine to form two compounds XY2 and X3Y2. When

0.1 mole of XY2 weighs 10 g and 0.05 mole of X3Y 2 weighs 9 g, the atomic weights of X and
are

(a)
40, 30
(b)
60, 40

(NEET-II 2016)

What is the mass of the precipitate formed when 50 mL of 16.9% solution of AgNO3 is

mixed with 50 mL of 5.8% NaCl solution? (Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5)

(a) 3.5 g
(b) 7 g
(c) 14 g (d) 28 g

(2015)

If Avogadro number NA, is changed from 6.022 × 1023 mol–1 to 6.022 × 1020 mol–1, this would change

the mass of one mole of carbon

the ratio of chemical species to each other in a balanced equation

the ratio of elements to each other in a compound

the definition of mass in units of grams. (2015)

The number of water molecules is maximum in

1.8 gram of water

18 gram of water

18 moles of water

(d) 18 molecules of water.
(2015)

A mixture of gases contains H2 and O2 gases in the ratio of 1 : 4 (w/w). What is the molar ratio of the two gases in the mixture?

16 : 1 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1 (2015, Cancelled)

Equal masses of H2, O2 and methane have been taken in a container of volume V at temperature 27 °C in identical conditions. The ratio of the volumes of gases H2 : O2 : methane would be

(a)
8 : 16 : 1
(b)
16 : 8 : 1

When 22.4 litres of H2(g) is mixed with 11.2 litres of Cl2(g), each at S.T.P, the moles of HCl(g)
formed is equal to
(a)
1 mol of HCl(g)
(b)
2 mol of HCl(g)

(2014)

1.0 g of magnesium is burnt with 0.56 g O2 in

a closed vessel. Which reactant is left in excess and how much? (At. wt. Mg = 24, O = 16)
(a)
Mg, 0.16
g
(b)
O2, 0.16 g

(2014)

6.02 × 1020 molecules of urea are present in 100 mL of its solution. The concentration of solution is

(a)
0.001 M
(b)
0.1 M

(NEET 2013)

10. In an experiment it showed that 10 mL of 0.05 M solution of chloride required 10 mL of 0.1 M solution of AgNO3, which of the following will be the formula of the chloride (X stands for the symbol of the element other than chlorine)

X2Cl2 (b) XCl2 (c) XCl4 (d) X2Cl (Karnataka NEET 2013)

Which has the maximum number of molecules among the following?
(a)
44 g CO2
(b)
48 g O3

(Mains 2011)

12. The number of atoms in 0.1 mol of a triatomic
gas is (NA = 6.02 × 1023 mol–1)

(a)
6.026 × 1022
(b)
1.806 × 1023

(2010)

13. 25.3 g of sodium carbonate, Na2CO3 is dissolved in enough water to make 250 mL of solution. If sodium carbonate dissociates completely, molar concentration of sodium ion,

Na+ and carbonate ions, CO32– are respectively (Molar mass of Na2CO3 = 106 g mol–1)
0.955 M and 1.910 M

1.910 M and 0.955 M

1.90 M and 1.910 M

(d) 0.477 M and 0.477 M
(2010)

14. 10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be

(a)
3 mol
(b)
4 mol

15. What volume of oxygen gas (O2) measured at 0°C and 1 atm, is needed to burn completely

L of propane gas (C3H8) measured under the same conditions?

(a)5 L(b) 10 L(c) 7 L(d) 6 L

(2008)

16. How many moles of lead (II) chloride will be formed from a reaction between 6.5 g of PbO and 3.2 g HCl?

0.011 (b) 0.029 (c) 0.044 (d) 0.333

(2008)

17. An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C, 38.71% and H, 9.67%. The empirical formula of the compound would be

(a)
CHO
(b)
CH4O

An element, X has the following isotopic composition:

200X : 90% 199X : 8.0% 202X : 2.0% The weighted average atomic mass of the naturally occurring element X is closest to

(a)
201 amu
(b)
202 amu

The maximum number of molecules is present in

15 L of H2 gas at STP
5 L of N2 gas at STP
0.5 g of H2 gas
(d) 10 g of O2 gas.
(2004)

Which has maximum molecules?
(a)
7 g N2
(b)
2 g H2

Percentage of Se in peroxidase anhydrous enzyme is 0.5% by weight (at. wt. = 78.4) then minimum molecular weight of peroxidase anhydrous enzyme is

(a)
1.568 × 104
(b)
1.568 × 103

(2001)

Molarity of liquid HCl, if density of solution is

1.17 g/cc is

(a)
36.5
(b)
18.25

Specific volume of cylindrical virus particle is 6.02 × 10–2 cc/g whose radius and length are

Å and 10 Å respectively. If NA = 6.02 × 1023, find molecular weight of virus.
(a)15.4 kg/mol(b) 1.54 × 104 kg/mol
(c) 3.08 × 104 kg/mol (d) 3.08 × 103 kg/mol (2001)

In quantitative analysis of second group in laboratory, H2S gas is passed in acidic medium for precipitation. When Cu2+ and Cd2+ react with KCN, then for product, true statement is
K2[Cu(CN)4] more soluble
K2[Cd(CN)4] less stable
K3[Cu(CN)2] less stable
(d) K2[Cd(CN)3] more stable.
(2000)

Volume of CO2 obtained by the complete
decomposition of 9.85 g of BaCO3 is

(a)
2.24 L
(b)
1.12 L

Oxidation numbers of A, B, C are +2, +5 and –2 respectively. Possible formula of compound is
(a)
A2(BC2)2
(b)
A3(BC4)2

(2000)

The number of atoms in 4.25 g of NH3 is approximately
(a)
4 × 1023
(b)
2
× 2023

Given the numbers: 161 cm, 0.161 cm, 0.0161 cm. The number of significant figures for the three numbers is
3, 3 and 4 respectively

3, 4 and 4 respectively

3, 4 and 5 respectively

(d) 3, 3 and 3 respectively.
(1998)

(d) As+ (1998)

(d) 2 (1998)

Some Basic Concepts of Chemistry
3

Haemoglobin contains 0.334% of iron by weight. The molecular weight of haemoglobin is approximately 67200. The number of iron atoms (Atomic weight of Fe is 56) present in

one molecule of haemoglobin is

(a)4(b) 6(c) 3

In the reaction,
4NH3(g) + 5O2(g) ® 4NO(g) + 6H2O(l)
when 1 mole of ammonia and 1 mole of O2 are made to react to completion :

All the oxygen will be consumed.

1.0 mole of NO will be produced.
1.0 mole of H2O is produced.
All the ammonia will be consumed. (1998)

Among the following which one is not paramagnetic? [Atomic numbers; Be = 4, Ne = 10, As = 33, Cl = 17]
(a)Ne2+(b) Be+(c) Cl–

0.24 g of a volatile gas, upon vaporisation, gives

45 mL vapour at NTP. What will be the vapour density of the substance? (Density of H2 = 0.089)
95.93 (b) 59.93 (c) 95.39 (d) 5.993

(1996)

The amount of zinc required to produce 224 mL of H2 at STP on treatment with dilute H2SO4
will be

(a) 65 g
(b) 0.065 g (c) 0.65 g (d) 6.5 g

(1996)

The dimensions of pressure are the same as that of

force per unit volume

energy per unit volume

force

(d) energy.
(1995)

The number of moles of oxygen in one litre of

air containing 21% oxygen by volume, under standard conditions, is

(a)
0.0093 mol
(b)
2.10 mol

(1995)

The total number of valence electrons in 4.2 g
of N3– ion is (NA is the Avogadro’s number)
(a)
2.1
NA
(b)
4.2
NA

A 5 molar solution of H2SO4 is diluted from 1

litre to a volume of 10 litres, the normality of the solution will be

(a)
1 N
(b)
0.1 N

The number of gram molecules of oxygen in
6.02 × 1024 CO molecules is
(a)10 g molecules(b) 5 g molecules

(1990)

Boron has two stable isotopes, 10B(19%) and 11B(81%). Calculate average at. wt. of boron in the periodic table

(a)
10.8
(b)
10.2

The molecular weight of O2 and SO2 are 32 and 64 respectively. At 15°C and 150 mmHg pressure, one litre of O2 contains ‘N’ molecules. The number of molecules in two litres of SO2 under the same conditions of temperature and pressure will be

(a)
N/2
(b)
N

A metal oxide has the formula Z2O3. It can be reduced by hydrogen to give free metal and water. 0.1596 g of the metal oxide requires 6 mg

of hydrogen for complete reduction. The atomic weight of the metal is

(a)
27.9
(b)
159.6

Ratio of Cp and CV of a gas ‘X’ is 1.4. The

number of atoms of the gas ‘X’ present in 11.2 litres of it at NTP will be
(a)
6.02 × 1023
(b)
1.2 × 1023

(1989)

What is the weight of oxygen required for the

complete combustion of 2.8 kg of ethylene?

(a) 2.8 kg (b) 6.4 kg (c) 9.6 kg (d) 96 kg (1989)

The number of oxygen atoms in 4.4 g of CO2 is
(a)
1.2 × 1023
(b)
6 × 1022

(1989)

45. At S.T.P. the density of CCl4 vapour in g/L will be nearest to
(a)6.87(b) 3.42(c) 10.26(d) 4.57

(1988)

One litre hard water contains 12.00 mg Mg2+. Milli-equivalents of washing soda required to

remove its hardness is
(a)
1
(b)
12.16

(1988)

47. 1 cc N2O at NTP contains

2241.8 u10 22 atoms

224006.02 u10 23 molecules

1.32224 u10 23electrons

(d) All the above.
(1988)

Answer Key

1.
(a)
2.
(b)
3.
(a)
4.
(c)
5.
(d)
6.
(c)
7.
(a)
8.
(a)
9.
(d)
10.
(b)

11.
(c)
12.
(b)
13.
(b)
14.
(b)
15.
(a)
16.
(b)
17.
(c)
18.
(d)
19.
(a)
20.
(b)

21.
(a)
22.
(c)
23.
(a)
24.
(c)
25.
(b)
26.
(b)
27.
(d)
28.
(d)
29.
(a)
30.
(a)

31.
(c)
32.
(b)
33.
(c)
34.
(b)
35.
(a)
36.
(c)
37.
(a)
38.
(b)
39.
(a)
40.
(c)

41.
(d)
42.
(a)
43.
(c)
44.
(a)
45.
(a)
46.
(a)
47.
(d)

Some Basic Concepts of Chemistry

(a) : Let atomic weight of element X is x and that of element Y is y.

For XY2, ⇒ x + 2y =

...(i) For X3Y2,
⇒

...(ii) On solving equations (i) and (ii), we get y = 30 x + 2(30) = 100 ⇒ x = 100 – 60 = 40
(b) : 16.9% solution of AgNO3 means 16.9 g of AgNO3 in 100 mL of solution.

9 g of AgNO3 in 100 mL solution ≡ 8.45 g of AgNO3 in 50 mL solution.

Similarly, 5.8% of NaCl in 100 mL solution º 2.9 g of NaCl in 50 mL solution.
The reaction can be represented as :
Initial
AgNO3
+ NaCl
AgCl + NaNO3

8.45/170
2.9/58.5
0
0

mole
= 0.049
= 0.049

Final moles0
0
0.049
0.049

\ Mass of AgCl precipitated = 0.049 × 143.3 = 7.02 » 7 g

(a) : Mass of 1 mol (6.022 × 1023 atoms) of carbon = 12 g

If Avogadro number is changed to 6.022 × 1020 atoms then mass of 1 mol of carbon
12 u 6.022 u1020
12 u10 3g

6.022 u1023

4. (c) : 1.8 gram of water =
6.023 u1023
u1.8

= 6.023 × 1022 molecules
18 gram of water = 6.023 × 1023 molecules
18 moles of water = 18 × 6.023 × 1023 molecules
1

5.(d) : Number of moles of H2 = 2

4
Number of moles of O2 = 32

Hence, molar ratio = 12 : 324 = 4 : 1

(c) : According to Avogadro’s hypothesis, ratio of the volumes of gases will be equal to the ratio of their no. of moles.

So, no. of moles =

Mass

Mol. mass

n
w
;
n

w
; n

H 2

O 2

CH4

w
w

So, the ratio is

or 16 :1: 2.

(a) : 1 mole ≡ 22.4 litres at S.T.P.

22.4
1 mol ;
n

11.2

0.5 mol

22.4

Cl2

Reaction is as,

H2(g)
+
Cl2(g)

2HCl(g)

Initial
1 mol

0.5 mol

Final
(1 – 0.5)
(0.5 – 0.5)
2
× 0.5

= 0.5 mol
= 0 mol

1
mol

Here, Cl2 is limiting reagent. So, 1 mole of HCl(g) is formed.

8. (a) : n

1
= 0.0416 moles

n
0.56
= 0.0175 moles

The balanced equation is

Mg
+

O2
MgO

Initial
0.0416 moles
0.0175 moles
0

Final
(0.0416 – 2 × 0.0175)
0
2 × 0.0175

0.0066 moles (O 2 is limiting reagent.)
Mass of Mg left in excess = 0.0066 × 24 = 0.16 g

9. (d) : Moles of urea =
×
= 0.001

Concentration of solution = × 1000 = 0.01 M

(b) : Millimoles of solution of chloride

= 0.05 × 10 = 0.5
Millimoles of AgNO3 solution = 10 × 0.1 = 1

So, the millimoles of AgNO3 are double than the chloride solution.
XCl2 + 2AgNO3 → 2AgCl + X(NO3)2
11. (c) : 8 g H2 has 4 moles while the others has
1 mole each.
12.
(b) : No. of atoms = NA
× No. of moles × 3

= 6.023 × 1023 × 0.1 × 3 = 1.806 × 1023

13.
(b) : Given that molar mass of Na2CO3 = 106 g

∴
Molarity of solution =

25.3
×1000

106
× 250

0.9547 M = 0.955 M Na2CO3 → 2Na+ + CO23–

[Na+] = 2[Na2CO3] = 2 × 0.955 = 1.910 M [CO2–3] = [Na2CO3] = 0.955 M

(b) : H2 + 1/2O2 → H2O
2 g
16 g
18 g

1 mol
0.5
1 mol

10 g of H2 = 5 mol and 64 g of O 2 = 2 mol ∴ In this reaction, oxygen is the limiting reagent so
amount of H2O produced depends on that of O2.
Since 0.5 mol of O2 gives 1 mol H2O

∴ 2 mol of O2 will give 4 mol H2O 15. (a) :

According to the above equation
1 vol. or 1 litre of propane requires to 5 vol.
or 5 litre of O2 to burn completely.
16. (b) : PbO+2HCl → PbCl2 + H2O
n mol2n moln mol
mol 3.2 mol
22436.5

0.029 mol0.087 mol

Formation of moles of lead (II) chloride depends upon the no. of moles of PbO which acts as a limiting factor here. So, no. of moles of PbCl2 formed will be equal to the no. of moles of PbO i.e. 0.029.

17. (c) :

Element
%
Atomic

mole

simple

mass

ratio

C
38.71
12

38.71
= 3.22

3.22
=1

3.22

H
9.67
1

9.67
= 9.67

9.67

= 3

3.22

O
51.62
16

51.62

= 3.22

3.22
=1

3.22

Hence empirical formula of the compound would be CH3O.

(d) : Average isotopic mass of X

=
200 × 90 + 199 × 8 + 202 × 2

90 + 8 + 2

= 18000 + 1592 + 404 = 199.96 a.m.u. ≈ 200 a.m.u. 100

(a) : At STP, 22.4 L H2 = 6.023 × 1023 molecules

15 L H =

6.023× 1023 ×15

= 4.033×1023

22.4

5 L N

6.023 ×10 23 × 5
=
1.344 ×1023

22.4

2 g H2 = 6.023 × 1023

0.5 g H2 =6.023 × 10 23 × 0.5 = 1.505 ×10 23
2

32 g O2 = 6.023 × 1023
g of O = 6.023× 1023 ×10 = 1.882×1023
232

20.(b) : 1 mole of any element contain 6.023 × 1023 number of molecules.

1 g mole of O2 = 32 g O2 ⇒ 16 g of O2 = 0.5 g mole O2
1 g mole of N2 = 28 g N2
⇒7 g N2 = 0.25 g mole N2
1 g mole of H2 = 2 g H2
⇒2 g H2 = 1.0 g mole H2
1 g mole NO2 = 14 + 16 × 2 = 46 ⇒ 16 g of NO2 = 0.35 mole NO2
2 g H2 (1 g mole H2) contain maximum molecules.

(a) : In peroxidase anhydrous enzyme 0.5% Se is present means, 0.5 g Se is present in 100 g of enzyme. In a molecule of enzyme one Se atom must be present. Hence 78.4 g Se will be present in

1000.5 × 78.4 = 1.568 ×104

1 cc. solution contains 1.17 g of HCl

∴ Molarity =

1.17 ×1000
= 32.05

36.5 ×1

(a) : Specific volume (vol. of 1 g) cylindrical virus particle = 6.02 × 10–2 cc/g

Radius of virus, r = 7 Å = 7 × 10–8 cm Volume of virus = πr2l

227 × (7 × 10 − 8 ) 2 × 10 ×10−8 = 154 × 10–23 cc
Volume
wt. of one virus particle = Specific volume

⇒
1 5 4 × 1 0
− 2 3
g

6 .0 2
× 1 0 −
2

Molecular wt. of virus = wt. of NA particle
154 ×10−23 × 6.02 ×10 −23 g/mol.
×10−2

15400 g/mol = 15.4 kg/mol

(c) : K3[Cu(CN)2] = 3(+1) + x + 2(–1) = 0 ⇒ x = –1
As the oxidation no. of ‘Cu’ is –1 (–ve), so this complex is unstable and is not formed.

(b) : BaCO3 → BaO + CO2

197 ⋅ 34 g → 22 ⋅ 4 L at N.T.P.

9 ⋅85 g →

22 ⋅ 4
× 9 ⋅85 = 1⋅118 L

197 ⋅ 34

Some Basic Concepts of Chemistry
7

9 ⋅85 g BaCO3 will produce 1.118 L CO2 at N.T.P. on the complete decomposition.
(b) : In A3(BC4)2, (+2) × 3 + 2[+5 + 4(-2)]
+ 6 + 10 − 16 = 0
Hence in the compound A3(BC4)2, the oxidation no.

of ‘A’, ‘B’ and ‘C’ are +2, +5 and −2 respectively.

(d) : No. of molecules in 4.25 g NH3

=
4.25
× 6.023 × 1023
= 2.5 × 6.023 × 1022

Number of atoms in 4.25 g NH3
4 × 2.5 × 6.023 × 1022 = 6.023 × 1023

(d) : Zeros placed left to the number are never significant, therefore the no. of significant figures for the numbers.

cm = 0.161 cm and 0.0161 cm are same, i.e. 3

(a) : Quantity of iron in one molecule
67200100 × 0.334 = 224.45 amu

No. of iron atoms in one molecule of haemoglobin
224.45
=56= 4

(a) : 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l) 4 mole + 5 mole → 4 mole + 6 mole

1 mole of NH3 requires = 54 = 1.25 mole of oxygen

while 1 mole of O2 requires = 54 = 0.8 mole of NH3. As there is 1 mole of NH3 and 1 mole of O2, so all oxygen will be consumed.

Cl–(18) ⇒ 1s22s22p63s23p6

As + (32) ⇒1s 2 2 s 2 2 p 6 3s 2 3 p 6 4 s 2 3d 10 4 p1x 4 p1y

Cl– is not paramagnetic , as it has no unpaired electron.

(b) : Weight of gas = 0.24 g, Volume of gas =
45 mL = 0.045 litre and density of H2 = 0.089.
We know that weight of 45 mL of H2 =
Density × Volume = 0.089 × 0.045 = 4.005 × 10−3 g
Therefore vapour density
Weight of certain volume of substance
Weight of same volume of hydrogen

=
0.24
= 59.93

4.005 ×10−3

33. (c) : Zn

+ H2SO4 → ZnSO4 + H2

(65 g)
(22400 mL)

Since 65 g of zinc reacts to liberate 22400 mL of H2 at STP, therefore amount of zinc needed to produce 224 mL of H2 at STP

=
65
× 224 = 0.65 g

22400

34. (b) : Pressure =
Force

Area

MLT
−2

Therefore dimensions of pressure =

= ML−1T–2

and dimensions of energy per unit volume

Energy
ML2 T−2
= ML−1T−2

Volume

(a) : Volume of oxygen in one litre of air
10021 ×1000 = 210 mL

210
Therefore no. of mol = 22400 = 0.0093 mol

(c) : Each nitrogen atom has 5 valence electrons, therefore total number of electrons in N3− ion is 16. Since the molecular mass of N3 is 42, therefore total number of electrons in 4.2 g of N3− ion

4.242 ×16 × NA = 1.6 NA

(a) : 5M H2SO4 = 10N H2SO4
N1V1 = N2V2 Þ 10 × 1 = N2 × 10 Þ N2 = 1N
(b) : Avogadro’s No., NA = 6.02 × 1023 molecules. \ 6.02 × 1024 CO molecules = 10 moles CO

10 g atoms of O = 5 g molecules of O2

(a) : Average atomic mass

19 u 10 81 u11
10.81

100

(d) : Z2O3 + 3H2 ® 2Z + 3H2O
Valency of metal in Z2O3 = 3
0.1596 g of Z2O3 react with 6 mg of H2.

[1 mg = 0.001 g = 10–3g]

\ 1 g of H2 react with
0.1596
= 26.6 g of Z2O3

0.006

\ Eq. wt. of Z2O3 = 26.6

Now, Eq. wt. of Z + Eq. wt. of O = Eq. wt. of Z + 8 = 26.6

Eq. wt. of Z = 26.6 – 8 = 18.6 \ At. wt. of Z = 18.6 × 3 = 55.8
Atomic wt. º
«Eq. wt =»
¬Valency of metal¼

(a) : Here, Cp/CV = 1.4, which shows that the gas is diatomic.

22.4 L at NTP = 6.02 × 1023 molecules

\ 11.2 L at NTP = 3.01 × 1023 molecules
Since gas is diatomic.

\ 11.2 L at NTP = 6.02 × 1023 atom

28 g 96 g

2.8 kg C H
=
96 g
u 2.8 kg

2
4
28 g

9628 u 2.8 u103 g = 9.6 × 103 g = 9.6 kg

(a) : 1 mol of CO2 = 44 g of CO2

\ 4·4 g CO2 = 0.1 mol CO2 = 6 × 1022 molecules

[Since, 1 mole CO2 = 6 × 1023 molecules]

2 × 6 × 1022 atoms of O = 1.2 × 1023 atoms of O

(a) : 1 mol CCl4 vapour = 12 + 4 × 35.5
154 g = 22.4 L

\ Density of CCl4 vapour
154
g L–1

22.4

= 6.875 g L–1

(a) : Mg2+ + Na2CO3 → MgCO3 + 2Na+

1g eq.1g eq.
1g eq. of Mg2+ = 12g of Mg2+ = 12000 mg

Now, 1000 millieq. of Na2CO3 = 12000 mg of Mg2+ \ 1 millieq. of Na2CO3 = 12 mg of Mg2+

(d) : As we know,

22400 cc of N2O contain 6.02 × 1023 molecules
\ 1 cc of N2O contain 6.02 u1023 molecules

22400

Since in N2O molecule there are 3 atoms
\ 1 cc N2O
3 u 6.02 u1023
atoms

22400

1.8 u1022
atoms

224

No. of electrons in a molecule of N2O = 7 + 7 + 8 = 22
Hence, no. of electrons
6.02 u1023
u 22 electrons

22400

1.32
u1023 electrons

224

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