Basic Concepts of Chemistry Practice Questions - DPP for JEE

 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