NEET Exam > NEET Notes > Chemistry Class 11 > DPP for NEET: Daily Practice Problems, Ch: Some Basic Concepts of Chemistry (Solutions)
Some Basic Concepts of Chemistry Practice Questions - DPP for NEET
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Page 1
1. (c)
According to law of conservation of mass "mass is neither created nor
destroyed during a chemical change"
? Mass of the reactants = Mass of products x + 4.9 = 6 +
1.825
or x = 2.925 g
2. (b) Moles of urea present in 100 ml of sol.=
?
[ M = Moles of solute present in 1L of solution]
3. (d) Relative atomic mass =
or
Now if we use
1
/
6
in place of
1
/
12
the formula becomes
? .
4. (d) 1 Mole of Mg
3
(PO
4
)
2
contains 8 mole of oxygen atoms
? 8 mole of oxygen atoms = 1 mole of Mg
3
(PO
4
)
2
?
mole of
Mg
3
(PO
4
)
2
0.25 mole of oxygen atom mole of Mg
3
(PO
4
)
2
mole of Mg
3
(PO
4
)
2
5. (b) From
Page 2
1. (c)
According to law of conservation of mass "mass is neither created nor
destroyed during a chemical change"
? Mass of the reactants = Mass of products x + 4.9 = 6 +
1.825
or x = 2.925 g
2. (b) Moles of urea present in 100 ml of sol.=
?
[ M = Moles of solute present in 1L of solution]
3. (d) Relative atomic mass =
or
Now if we use
1
/
6
in place of
1
/
12
the formula becomes
? .
4. (d) 1 Mole of Mg
3
(PO
4
)
2
contains 8 mole of oxygen atoms
? 8 mole of oxygen atoms = 1 mole of Mg
3
(PO
4
)
2
?
mole of
Mg
3
(PO
4
)
2
0.25 mole of oxygen atom mole of Mg
3
(PO
4
)
2
mole of Mg
3
(PO
4
)
2
5. (b) From
V
1
=
6. (d) Weight of Iron in 67200 =
Number of atoms of Iron =
7. (a) 2Al(s) + 6HCl(aq) 2Al
3+
(aq) + 6Cl
–
(aq) + 3H
2
(g)
6 moles of HCl produces = 3 moles of H
2
= 3 × 22.4 L of H
2
at
S.T.P
1 mole of HCl produces = L of H
2
at S.T.P = 11.2 L of
H
2
at
STP
8. (a) 95% H
2
SO
4
by weight means 100g H
2
SO
4
solution contains 95g
H
2
SO
4
by mass.
Molar mass of H
2
SO
4
= 98g mol
–1
Moles in 95g = = 0.969 mole
Volume of 100g = 54.52 cm
3
= 54.52
× 10
–3
L
9. (c) 50 mL of 16.9% solution of AgNO
3
= 8.45 g of Ag NO
3
n
mole
= = 0.0497 moles
50 ml of 5.8% solution of NaCl contain
Page 3
1. (c)
According to law of conservation of mass "mass is neither created nor
destroyed during a chemical change"
? Mass of the reactants = Mass of products x + 4.9 = 6 +
1.825
or x = 2.925 g
2. (b) Moles of urea present in 100 ml of sol.=
?
[ M = Moles of solute present in 1L of solution]
3. (d) Relative atomic mass =
or
Now if we use
1
/
6
in place of
1
/
12
the formula becomes
? .
4. (d) 1 Mole of Mg
3
(PO
4
)
2
contains 8 mole of oxygen atoms
? 8 mole of oxygen atoms = 1 mole of Mg
3
(PO
4
)
2
?
mole of
Mg
3
(PO
4
)
2
0.25 mole of oxygen atom mole of Mg
3
(PO
4
)
2
mole of Mg
3
(PO
4
)
2
5. (b) From
V
1
=
6. (d) Weight of Iron in 67200 =
Number of atoms of Iron =
7. (a) 2Al(s) + 6HCl(aq) 2Al
3+
(aq) + 6Cl
–
(aq) + 3H
2
(g)
6 moles of HCl produces = 3 moles of H
2
= 3 × 22.4 L of H
2
at
S.T.P
1 mole of HCl produces = L of H
2
at S.T.P = 11.2 L of
H
2
at
STP
8. (a) 95% H
2
SO
4
by weight means 100g H
2
SO
4
solution contains 95g
H
2
SO
4
by mass.
Molar mass of H
2
SO
4
= 98g mol
–1
Moles in 95g = = 0.969 mole
Volume of 100g = 54.52 cm
3
= 54.52
× 10
–3
L
9. (c) 50 mL of 16.9% solution of AgNO
3
= 8.45 g of Ag NO
3
n
mole
= = 0.0497 moles
50 ml of 5.8% solution of NaCl contain
NaCl =
n
NaCl
= = 0.0495 moles
AgNO
3
+ NaCl ? AgCl? + Na + Cl
1 mole 1 mole 1 mole
? 0.049 mole 0.049 mole 0.049 mole of AgCl
n =
w = (n
AgCl
) × Molecular Mass
= (0.049) × (107.8 + 35.5)
= 7.02 g
10. (b) Number of valence electrons in a ion = 1
Now, 1 mol or 42 g of has = 6.023 × 10
23
ions
So, 42 g of has 6.023 × 4 × 10
23
valence e
–
1 g of has valence e
–
4.2 g of has valence e
–
i.e.,
0.1 N
A
valence e
–
.
11. (c) 74.75% of chlorine means 74.75g chlorine is present in 100g of
metal chloride.
Weight of metal = 100g – 74.75g = 25.25g
Equivalent weight = 12
Valency of metal
? Formula of compound = MCl
4
12. (d) Q 18 gm, H
2
O contains = 2 gm H
Page 4
1. (c)
According to law of conservation of mass "mass is neither created nor
destroyed during a chemical change"
? Mass of the reactants = Mass of products x + 4.9 = 6 +
1.825
or x = 2.925 g
2. (b) Moles of urea present in 100 ml of sol.=
?
[ M = Moles of solute present in 1L of solution]
3. (d) Relative atomic mass =
or
Now if we use
1
/
6
in place of
1
/
12
the formula becomes
? .
4. (d) 1 Mole of Mg
3
(PO
4
)
2
contains 8 mole of oxygen atoms
? 8 mole of oxygen atoms = 1 mole of Mg
3
(PO
4
)
2
?
mole of
Mg
3
(PO
4
)
2
0.25 mole of oxygen atom mole of Mg
3
(PO
4
)
2
mole of Mg
3
(PO
4
)
2
5. (b) From
V
1
=
6. (d) Weight of Iron in 67200 =
Number of atoms of Iron =
7. (a) 2Al(s) + 6HCl(aq) 2Al
3+
(aq) + 6Cl
–
(aq) + 3H
2
(g)
6 moles of HCl produces = 3 moles of H
2
= 3 × 22.4 L of H
2
at
S.T.P
1 mole of HCl produces = L of H
2
at S.T.P = 11.2 L of
H
2
at
STP
8. (a) 95% H
2
SO
4
by weight means 100g H
2
SO
4
solution contains 95g
H
2
SO
4
by mass.
Molar mass of H
2
SO
4
= 98g mol
–1
Moles in 95g = = 0.969 mole
Volume of 100g = 54.52 cm
3
= 54.52
× 10
–3
L
9. (c) 50 mL of 16.9% solution of AgNO
3
= 8.45 g of Ag NO
3
n
mole
= = 0.0497 moles
50 ml of 5.8% solution of NaCl contain
NaCl =
n
NaCl
= = 0.0495 moles
AgNO
3
+ NaCl ? AgCl? + Na + Cl
1 mole 1 mole 1 mole
? 0.049 mole 0.049 mole 0.049 mole of AgCl
n =
w = (n
AgCl
) × Molecular Mass
= (0.049) × (107.8 + 35.5)
= 7.02 g
10. (b) Number of valence electrons in a ion = 1
Now, 1 mol or 42 g of has = 6.023 × 10
23
ions
So, 42 g of has 6.023 × 4 × 10
23
valence e
–
1 g of has valence e
–
4.2 g of has valence e
–
i.e.,
0.1 N
A
valence e
–
.
11. (c) 74.75% of chlorine means 74.75g chlorine is present in 100g of
metal chloride.
Weight of metal = 100g – 74.75g = 25.25g
Equivalent weight = 12
Valency of metal
? Formula of compound = MCl
4
12. (d) Q 18 gm, H
2
O contains = 2 gm H
? 0.72 gm H
2
O contains
=
Q 44 gm CO
2
contains = 12 gm C
? 3.08 gm CO
2
contains
=
? C : H =
= 0.07 : 0.08 = 7 : 8
? Empirical formula = C
7
H
8
13. (a)
Na
2
CO
3
+ NaHCO
3
+ NaCl + HCl
(excess)
Thus, on complete reaction with HCl, 1kg of washing soda will evolve 9
mol of CO
2
.
14. (a) 2.6 has two significant figures.
0.260 has three significant figures.
0.002600 has four significant figures.
2.6000 has five significant figures.
15. (b) Given
mass of solute (w) = 120 g
mass of solvent (w) = 1000 g
Mol. mass of solute = 60 g
density of solution = 1.12 g/ ml
From the given data,
Page 5
1. (c)
According to law of conservation of mass "mass is neither created nor
destroyed during a chemical change"
? Mass of the reactants = Mass of products x + 4.9 = 6 +
1.825
or x = 2.925 g
2. (b) Moles of urea present in 100 ml of sol.=
?
[ M = Moles of solute present in 1L of solution]
3. (d) Relative atomic mass =
or
Now if we use
1
/
6
in place of
1
/
12
the formula becomes
? .
4. (d) 1 Mole of Mg
3
(PO
4
)
2
contains 8 mole of oxygen atoms
? 8 mole of oxygen atoms = 1 mole of Mg
3
(PO
4
)
2
?
mole of
Mg
3
(PO
4
)
2
0.25 mole of oxygen atom mole of Mg
3
(PO
4
)
2
mole of Mg
3
(PO
4
)
2
5. (b) From
V
1
=
6. (d) Weight of Iron in 67200 =
Number of atoms of Iron =
7. (a) 2Al(s) + 6HCl(aq) 2Al
3+
(aq) + 6Cl
–
(aq) + 3H
2
(g)
6 moles of HCl produces = 3 moles of H
2
= 3 × 22.4 L of H
2
at
S.T.P
1 mole of HCl produces = L of H
2
at S.T.P = 11.2 L of
H
2
at
STP
8. (a) 95% H
2
SO
4
by weight means 100g H
2
SO
4
solution contains 95g
H
2
SO
4
by mass.
Molar mass of H
2
SO
4
= 98g mol
–1
Moles in 95g = = 0.969 mole
Volume of 100g = 54.52 cm
3
= 54.52
× 10
–3
L
9. (c) 50 mL of 16.9% solution of AgNO
3
= 8.45 g of Ag NO
3
n
mole
= = 0.0497 moles
50 ml of 5.8% solution of NaCl contain
NaCl =
n
NaCl
= = 0.0495 moles
AgNO
3
+ NaCl ? AgCl? + Na + Cl
1 mole 1 mole 1 mole
? 0.049 mole 0.049 mole 0.049 mole of AgCl
n =
w = (n
AgCl
) × Molecular Mass
= (0.049) × (107.8 + 35.5)
= 7.02 g
10. (b) Number of valence electrons in a ion = 1
Now, 1 mol or 42 g of has = 6.023 × 10
23
ions
So, 42 g of has 6.023 × 4 × 10
23
valence e
–
1 g of has valence e
–
4.2 g of has valence e
–
i.e.,
0.1 N
A
valence e
–
.
11. (c) 74.75% of chlorine means 74.75g chlorine is present in 100g of
metal chloride.
Weight of metal = 100g – 74.75g = 25.25g
Equivalent weight = 12
Valency of metal
? Formula of compound = MCl
4
12. (d) Q 18 gm, H
2
O contains = 2 gm H
? 0.72 gm H
2
O contains
=
Q 44 gm CO
2
contains = 12 gm C
? 3.08 gm CO
2
contains
=
? C : H =
= 0.07 : 0.08 = 7 : 8
? Empirical formula = C
7
H
8
13. (a)
Na
2
CO
3
+ NaHCO
3
+ NaCl + HCl
(excess)
Thus, on complete reaction with HCl, 1kg of washing soda will evolve 9
mol of CO
2
.
14. (a) 2.6 has two significant figures.
0.260 has three significant figures.
0.002600 has four significant figures.
2.6000 has five significant figures.
15. (b) Given
mass of solute (w) = 120 g
mass of solvent (w) = 1000 g
Mol. mass of solute = 60 g
density of solution = 1.12 g/ ml
From the given data,
Mass of solution = 1000 + 120 = 1120 g
or
Volume of solution or = 1 litre
Now molarity (M) = =
16. (d) In an unknown compounds containing N and H
given % of H = 12.5%
? % of N = 100 – 12.5 = 87.5%
2 × vapour density = Mol. wt = mol wt. = 16 × 2 = 32.
Molecular formula = n × empirical formula mass
= 2
? Molecular formula of the compound will be = (NH
2
)
2
= N
2
H
4
17. (a)
18. (b) The required equation is
nascent
oxygen
[O] required for 1 mol. of Fe(C
2
O
4
) is 1.5, 5 [O] are obtained from 2
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| 1. What are the basic concepts of chemistry? |  |
Ans. The basic concepts of chemistry include elements, compounds, mixtures, atomic structure, chemical reactions, stoichiometry, and properties of matter.
| 2. What is the difference between an element and a compound? | |
Ans. An element is a substance that cannot be broken down into simpler substances by chemical means, while a compound is a substance composed of two or more different elements chemically combined in a fixed ratio.
| 3. How can atomic structure be described? | |
Ans. Atomic structure refers to the arrangement and properties of subatomic particles in an atom. It includes the nucleus (containing protons and neutrons) and the electron cloud (containing electrons in energy levels or orbitals).
| 4. What is stoichiometry in chemistry? | |
Ans. Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. It involves calculating the amount of substances involved, balancing chemical equations, and determining the limiting reactant.
| 5. What are the properties of matter? | |
Ans. The properties of matter include physical properties (such as color, density, melting point, boiling point) and chemical properties (such as reactivity, flammability, acidity). These properties help us understand and classify different types of matter.
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