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Basic Concepts of Chemistry Practice Questions - DPP for JEE
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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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