Some Basic Concepts of Chemistry Solved Examples | Chemistry for JEE Main & Advanced PDF Download

 Page 1


Solved Examples on Some Basic 
Concepts of Chemistry 
JEE Mains 
Q1. The density (in ?? ????
-?? ) of a ?? . ?????? sulphuric acid solution that is ???? %?? ?? ????
?? 
(molar mass = ???? ?? ?????? -?? ) by mass will be 
(a) 1.45 
(b) 1.64 
(c) 1.88 
(d) 1.22 
Ans: (d) Since molarity of solution is 3.60M. It means 3.6 moles of H
2
SO
4
 is present in its 
1 litre solution. 
Mass of 3.6 moles of H
2
SO
4
 
= Moles × Molecular mass = 3.6 × 98 g = 352 .8 g 
? 1000ml solution has 352 .8 g of H
2
SO
4
 
Given that 29 gof
2
SO
4
 is present in = 100 g of solution 
? 352 .8 g
g
 of H
2
SO
4
 is present in 
=
100
29
× 352 .8 g of solution = 1216 g of solution 
Density =
 Mass 
 Volume 
=
1216
1000
= 1.216 g/ml = 1.22 g/ml 
 
Q2. When 8 gms of oxygen reacts with magnesium then the amount of ?????? 
formed is 
(A) ???????? 
(B) ???????? 
(C) ???????? 
(D) ???????? 
Ans: (B) Mg +
1
2
O
2
? MgO 
0.25 mole 
0.5 mole 
Mass of MgO = 0.5 × (24 + 16) = 20gm 
 
Q3. One gram of the silver salt of an organic dibasic acid yields, on strong heating 
?? . ???????? ?? of silver. If the weight percentage of carbon in it 8 times the weight 
percentage of hydrogen and one half the weight percentage of oxygen, determine 
the molecular formula of the acid. [Atomic weight of ???? = ?????? ] 
(A) ?? ?? ?? ?? ?? ?? 
(B) ?? ?? ?? ?? ?? ?? 
(C) ?? ?? ?? ?? ?? ?? 
(D) ?? ?? ?? ????
?? ?? 
Page 2


Solved Examples on Some Basic 
Concepts of Chemistry 
JEE Mains 
Q1. The density (in ?? ????
-?? ) of a ?? . ?????? sulphuric acid solution that is ???? %?? ?? ????
?? 
(molar mass = ???? ?? ?????? -?? ) by mass will be 
(a) 1.45 
(b) 1.64 
(c) 1.88 
(d) 1.22 
Ans: (d) Since molarity of solution is 3.60M. It means 3.6 moles of H
2
SO
4
 is present in its 
1 litre solution. 
Mass of 3.6 moles of H
2
SO
4
 
= Moles × Molecular mass = 3.6 × 98 g = 352 .8 g 
? 1000ml solution has 352 .8 g of H
2
SO
4
 
Given that 29 gof
2
SO
4
 is present in = 100 g of solution 
? 352 .8 g
g
 of H
2
SO
4
 is present in 
=
100
29
× 352 .8 g of solution = 1216 g of solution 
Density =
 Mass 
 Volume 
=
1216
1000
= 1.216 g/ml = 1.22 g/ml 
 
Q2. When 8 gms of oxygen reacts with magnesium then the amount of ?????? 
formed is 
(A) ???????? 
(B) ???????? 
(C) ???????? 
(D) ???????? 
Ans: (B) Mg +
1
2
O
2
? MgO 
0.25 mole 
0.5 mole 
Mass of MgO = 0.5 × (24 + 16) = 20gm 
 
Q3. One gram of the silver salt of an organic dibasic acid yields, on strong heating 
?? . ???????? ?? of silver. If the weight percentage of carbon in it 8 times the weight 
percentage of hydrogen and one half the weight percentage of oxygen, determine 
the molecular formula of the acid. [Atomic weight of ???? = ?????? ] 
(A) ?? ?? ?? ?? ?? ?? 
(B) ?? ?? ?? ?? ?? ?? 
(C) ?? ?? ?? ?? ?? ?? 
(D) ?? ?? ?? ????
?? ?? 
Ans: (B) Let's diabasic acid is C
x
H
y
O
?? 
Weight of ?? =
?? (12)
?? 
 Weight = H =
?? ?? Weight of O =
z(16)
?? ?? (12) = 8 × ?? ? 3?? = 2?? ?? (12) =
1
2
× 16(?? ) ? 3?? = 2?? ?? = ?? =
3
2
?? ? Empirical formula 
2
H
3
O
3
Ag
2
(C
?? H
?? O
?? ) ? 2Ag
2
0.5934
108
 
Mole of salt =
0.5934
2×108
=
1
[216+(24+3+48)?? ]
 
216+ 75x = 364 
x ~ z 
So the formula would be = C
4
H
6
O
6
 
 
Q4. Mass of sucrose ?? ????
?? ????
?? ????
 produced by mixing ???????? of carbon, ???????? of 
hydrogen and 56 liter ?? ?? at ?? ?????? and ?????? ?? according to given reaction, is 
?? (?? ) + ?? ?? ( ?? ) + ?? ?? ( ?? ) ? ?? ????
?? ????
?? ????
( ?? ) 
(A) 138.5 
(B) 155.5 
(C) 172.5 
(D) 199.5 
Ans: (B) 12C (s) + 11H
2
( g) +
11
2
O
2
( g) ? C
12
H
22
O
11
( s) 
84
12
12
1
56
22.4
7 12 2.5
 
Here O
2
 is limiting reagent 
Moles of C
12
H
22
O
11
 formed =
2.5
11
× 2 =
5
11
 mole 
 Mass =
5
11
× [(12 × 12) + 22 + (11 × 16)] = 155 .45gm 
Page 3


Solved Examples on Some Basic 
Concepts of Chemistry 
JEE Mains 
Q1. The density (in ?? ????
-?? ) of a ?? . ?????? sulphuric acid solution that is ???? %?? ?? ????
?? 
(molar mass = ???? ?? ?????? -?? ) by mass will be 
(a) 1.45 
(b) 1.64 
(c) 1.88 
(d) 1.22 
Ans: (d) Since molarity of solution is 3.60M. It means 3.6 moles of H
2
SO
4
 is present in its 
1 litre solution. 
Mass of 3.6 moles of H
2
SO
4
 
= Moles × Molecular mass = 3.6 × 98 g = 352 .8 g 
? 1000ml solution has 352 .8 g of H
2
SO
4
 
Given that 29 gof
2
SO
4
 is present in = 100 g of solution 
? 352 .8 g
g
 of H
2
SO
4
 is present in 
=
100
29
× 352 .8 g of solution = 1216 g of solution 
Density =
 Mass 
 Volume 
=
1216
1000
= 1.216 g/ml = 1.22 g/ml 
 
Q2. When 8 gms of oxygen reacts with magnesium then the amount of ?????? 
formed is 
(A) ???????? 
(B) ???????? 
(C) ???????? 
(D) ???????? 
Ans: (B) Mg +
1
2
O
2
? MgO 
0.25 mole 
0.5 mole 
Mass of MgO = 0.5 × (24 + 16) = 20gm 
 
Q3. One gram of the silver salt of an organic dibasic acid yields, on strong heating 
?? . ???????? ?? of silver. If the weight percentage of carbon in it 8 times the weight 
percentage of hydrogen and one half the weight percentage of oxygen, determine 
the molecular formula of the acid. [Atomic weight of ???? = ?????? ] 
(A) ?? ?? ?? ?? ?? ?? 
(B) ?? ?? ?? ?? ?? ?? 
(C) ?? ?? ?? ?? ?? ?? 
(D) ?? ?? ?? ????
?? ?? 
Ans: (B) Let's diabasic acid is C
x
H
y
O
?? 
Weight of ?? =
?? (12)
?? 
 Weight = H =
?? ?? Weight of O =
z(16)
?? ?? (12) = 8 × ?? ? 3?? = 2?? ?? (12) =
1
2
× 16(?? ) ? 3?? = 2?? ?? = ?? =
3
2
?? ? Empirical formula 
2
H
3
O
3
Ag
2
(C
?? H
?? O
?? ) ? 2Ag
2
0.5934
108
 
Mole of salt =
0.5934
2×108
=
1
[216+(24+3+48)?? ]
 
216+ 75x = 364 
x ~ z 
So the formula would be = C
4
H
6
O
6
 
 
Q4. Mass of sucrose ?? ????
?? ????
?? ????
 produced by mixing ???????? of carbon, ???????? of 
hydrogen and 56 liter ?? ?? at ?? ?????? and ?????? ?? according to given reaction, is 
?? (?? ) + ?? ?? ( ?? ) + ?? ?? ( ?? ) ? ?? ????
?? ????
?? ????
( ?? ) 
(A) 138.5 
(B) 155.5 
(C) 172.5 
(D) 199.5 
Ans: (B) 12C (s) + 11H
2
( g) +
11
2
O
2
( g) ? C
12
H
22
O
11
( s) 
84
12
12
1
56
22.4
7 12 2.5
 
Here O
2
 is limiting reagent 
Moles of C
12
H
22
O
11
 formed =
2.5
11
× 2 =
5
11
 mole 
 Mass =
5
11
× [(12 × 12) + 22 + (11 × 16)] = 155 .45gm 
 
Q5. ?????????? of carbonate of an alkali metal or alkaline earth metal containing some 
inert impurities was made to react with excess ?????? solution. The liberated ????
?? 
occupied 12.315 lit. at ?? ?????? and ?????? ?? . The correct option is 
(A) Mass of impurity is ?????? and metal is Be 
(B) Mass of impurity is ?????? and metal is ???? 
(C) Mass of impurity is ?????? and metal is Be 
(D) Mass of impurity is ?????? and metal is ???? 
Ans: (B) M(CO
3
) ? CO
2
+ MO or 
M
2
(CO
3
) ? CO
2
+ M
2
O 
Mass of CO
3
= 12 + 48 = 60 
Mole of CO
2
=
12.315
(PV)
(RT ) =
12.315
1×(12.315)
× 0.0821× 
300 = 0.5 mole 
Mole of M(CO
3
) or M
2
CO
3
= 0.5 mole 
So, mass of CO
3
 in carbonate = 0.5 × 60 = 30 gram 
Checking all options one by one 
(B) is correct. 
Q7. The mass of ????
?? produced from 620 mixture of ?? ?? ?? ?? ?? ?? and ?? ?? ' prepared 
produce maximum energy is (combustion reaction is exothermic) 
(A) ?????? . ???????? 
(B) ?????? . ???????? 
(C) ?????????? 
(D) ?????????? 
Ans: (C) C
2
H
4
O
2
+ 2O
2
? 2CO
2
+ 2H
2
O 
x gm 620 - ?? gm 
11 11 
?? 60
 mole 
620 - ?? 32
 
 
To produce maximum energy C
2
H
4
O
2
 and O
2
 will be fully consumed. 
?? ?? 60
=
620 - ?? 32
×
1
2
64?? = 37200- 60?? ?? = 300gm
 
Weight of CO
2
= 2 ×
300×44
60
= 440gm 
Page 4


Solved Examples on Some Basic 
Concepts of Chemistry 
JEE Mains 
Q1. The density (in ?? ????
-?? ) of a ?? . ?????? sulphuric acid solution that is ???? %?? ?? ????
?? 
(molar mass = ???? ?? ?????? -?? ) by mass will be 
(a) 1.45 
(b) 1.64 
(c) 1.88 
(d) 1.22 
Ans: (d) Since molarity of solution is 3.60M. It means 3.6 moles of H
2
SO
4
 is present in its 
1 litre solution. 
Mass of 3.6 moles of H
2
SO
4
 
= Moles × Molecular mass = 3.6 × 98 g = 352 .8 g 
? 1000ml solution has 352 .8 g of H
2
SO
4
 
Given that 29 gof
2
SO
4
 is present in = 100 g of solution 
? 352 .8 g
g
 of H
2
SO
4
 is present in 
=
100
29
× 352 .8 g of solution = 1216 g of solution 
Density =
 Mass 
 Volume 
=
1216
1000
= 1.216 g/ml = 1.22 g/ml 
 
Q2. When 8 gms of oxygen reacts with magnesium then the amount of ?????? 
formed is 
(A) ???????? 
(B) ???????? 
(C) ???????? 
(D) ???????? 
Ans: (B) Mg +
1
2
O
2
? MgO 
0.25 mole 
0.5 mole 
Mass of MgO = 0.5 × (24 + 16) = 20gm 
 
Q3. One gram of the silver salt of an organic dibasic acid yields, on strong heating 
?? . ???????? ?? of silver. If the weight percentage of carbon in it 8 times the weight 
percentage of hydrogen and one half the weight percentage of oxygen, determine 
the molecular formula of the acid. [Atomic weight of ???? = ?????? ] 
(A) ?? ?? ?? ?? ?? ?? 
(B) ?? ?? ?? ?? ?? ?? 
(C) ?? ?? ?? ?? ?? ?? 
(D) ?? ?? ?? ????
?? ?? 
Ans: (B) Let's diabasic acid is C
x
H
y
O
?? 
Weight of ?? =
?? (12)
?? 
 Weight = H =
?? ?? Weight of O =
z(16)
?? ?? (12) = 8 × ?? ? 3?? = 2?? ?? (12) =
1
2
× 16(?? ) ? 3?? = 2?? ?? = ?? =
3
2
?? ? Empirical formula 
2
H
3
O
3
Ag
2
(C
?? H
?? O
?? ) ? 2Ag
2
0.5934
108
 
Mole of salt =
0.5934
2×108
=
1
[216+(24+3+48)?? ]
 
216+ 75x = 364 
x ~ z 
So the formula would be = C
4
H
6
O
6
 
 
Q4. Mass of sucrose ?? ????
?? ????
?? ????
 produced by mixing ???????? of carbon, ???????? of 
hydrogen and 56 liter ?? ?? at ?? ?????? and ?????? ?? according to given reaction, is 
?? (?? ) + ?? ?? ( ?? ) + ?? ?? ( ?? ) ? ?? ????
?? ????
?? ????
( ?? ) 
(A) 138.5 
(B) 155.5 
(C) 172.5 
(D) 199.5 
Ans: (B) 12C (s) + 11H
2
( g) +
11
2
O
2
( g) ? C
12
H
22
O
11
( s) 
84
12
12
1
56
22.4
7 12 2.5
 
Here O
2
 is limiting reagent 
Moles of C
12
H
22
O
11
 formed =
2.5
11
× 2 =
5
11
 mole 
 Mass =
5
11
× [(12 × 12) + 22 + (11 × 16)] = 155 .45gm 
 
Q5. ?????????? of carbonate of an alkali metal or alkaline earth metal containing some 
inert impurities was made to react with excess ?????? solution. The liberated ????
?? 
occupied 12.315 lit. at ?? ?????? and ?????? ?? . The correct option is 
(A) Mass of impurity is ?????? and metal is Be 
(B) Mass of impurity is ?????? and metal is ???? 
(C) Mass of impurity is ?????? and metal is Be 
(D) Mass of impurity is ?????? and metal is ???? 
Ans: (B) M(CO
3
) ? CO
2
+ MO or 
M
2
(CO
3
) ? CO
2
+ M
2
O 
Mass of CO
3
= 12 + 48 = 60 
Mole of CO
2
=
12.315
(PV)
(RT ) =
12.315
1×(12.315)
× 0.0821× 
300 = 0.5 mole 
Mole of M(CO
3
) or M
2
CO
3
= 0.5 mole 
So, mass of CO
3
 in carbonate = 0.5 × 60 = 30 gram 
Checking all options one by one 
(B) is correct. 
Q7. The mass of ????
?? produced from 620 mixture of ?? ?? ?? ?? ?? ?? and ?? ?? ' prepared 
produce maximum energy is (combustion reaction is exothermic) 
(A) ?????? . ???????? 
(B) ?????? . ???????? 
(C) ?????????? 
(D) ?????????? 
Ans: (C) C
2
H
4
O
2
+ 2O
2
? 2CO
2
+ 2H
2
O 
x gm 620 - ?? gm 
11 11 
?? 60
 mole 
620 - ?? 32
 
 
To produce maximum energy C
2
H
4
O
2
 and O
2
 will be fully consumed. 
?? ?? 60
=
620 - ?? 32
×
1
2
64?? = 37200- 60?? ?? = 300gm
 
Weight of CO
2
= 2 ×
300×44
60
= 440gm 
 
Q8. In the quantitative determination of nitrogen, ?? ?? gas liberated from ?? . ???????? of 
a sample of organic compound was collected over water. If the volume of ?? ?? gas 
collected was 
??????
???? ????
 at total pressure 860 ???????? at ?????? ?? , % by mass of nitrogen in 
the organic compound is 
[Aq. tension at ?????? ?? is ???? ???????? and ?? = ?? . ???? ?? atm ?????? -?? ?? -?? ] 
(A) 
????
?? % 
(B) 
?? ?? % 
(C) 
????
?? % 
(D) 
??????
?? % 
Ans: (A) (Organic compound) +H
2
O ? N
2
 0.42gm 
Moles of N
2
=
PV
RT
=
860
760
×
100
11
×
10
-3
0.08×250
 
=
86
167200
= 5.143× 10
-4
 
Mass of N
2
= 5.143× 10
-4
× 28 = 0.0144gm 
Fraction =
0.0144
0.42
= 0.034 =
10
3
% 
 
Q9. ?????? ???? of ?? . ?????????? and ?????? ???? of ?? . ???? ?? ?? ????
?? are mixed. The normality of the 
resulting mixture is 
(A) ?? . ?? ?? 
(B) ?? . ?? ?? 
(C) ?? . ?? ?? 
(D) ?? . ?? ?? 
Ans: (C) Moles of H
+
= (0.1)(0.3) + (0.2)(0.3) × 2 = 0.15 mole 
Normality =
0.15
500
× 1000 = 0.3 N 
 
Q10. The volume of water which should be added to ?????? ???? of ?? . ???????????? 
solution so as to get a solution of ?? . ???? is 
(A) ?????? ???? 
(B) ?????? ?? ?? 
(C) ?????? ???? 
(D) ?????? ???? 
Ans: (D) Moles of NaOH= (0.300)(0.5) = 0.15 moles 
For molarity = 0.2M =
0.15
 V
 
V = 750 mL 
Volume to be added = 750 - 300 = 450 mL 
Page 5


Solved Examples on Some Basic 
Concepts of Chemistry 
JEE Mains 
Q1. The density (in ?? ????
-?? ) of a ?? . ?????? sulphuric acid solution that is ???? %?? ?? ????
?? 
(molar mass = ???? ?? ?????? -?? ) by mass will be 
(a) 1.45 
(b) 1.64 
(c) 1.88 
(d) 1.22 
Ans: (d) Since molarity of solution is 3.60M. It means 3.6 moles of H
2
SO
4
 is present in its 
1 litre solution. 
Mass of 3.6 moles of H
2
SO
4
 
= Moles × Molecular mass = 3.6 × 98 g = 352 .8 g 
? 1000ml solution has 352 .8 g of H
2
SO
4
 
Given that 29 gof
2
SO
4
 is present in = 100 g of solution 
? 352 .8 g
g
 of H
2
SO
4
 is present in 
=
100
29
× 352 .8 g of solution = 1216 g of solution 
Density =
 Mass 
 Volume 
=
1216
1000
= 1.216 g/ml = 1.22 g/ml 
 
Q2. When 8 gms of oxygen reacts with magnesium then the amount of ?????? 
formed is 
(A) ???????? 
(B) ???????? 
(C) ???????? 
(D) ???????? 
Ans: (B) Mg +
1
2
O
2
? MgO 
0.25 mole 
0.5 mole 
Mass of MgO = 0.5 × (24 + 16) = 20gm 
 
Q3. One gram of the silver salt of an organic dibasic acid yields, on strong heating 
?? . ???????? ?? of silver. If the weight percentage of carbon in it 8 times the weight 
percentage of hydrogen and one half the weight percentage of oxygen, determine 
the molecular formula of the acid. [Atomic weight of ???? = ?????? ] 
(A) ?? ?? ?? ?? ?? ?? 
(B) ?? ?? ?? ?? ?? ?? 
(C) ?? ?? ?? ?? ?? ?? 
(D) ?? ?? ?? ????
?? ?? 
Ans: (B) Let's diabasic acid is C
x
H
y
O
?? 
Weight of ?? =
?? (12)
?? 
 Weight = H =
?? ?? Weight of O =
z(16)
?? ?? (12) = 8 × ?? ? 3?? = 2?? ?? (12) =
1
2
× 16(?? ) ? 3?? = 2?? ?? = ?? =
3
2
?? ? Empirical formula 
2
H
3
O
3
Ag
2
(C
?? H
?? O
?? ) ? 2Ag
2
0.5934
108
 
Mole of salt =
0.5934
2×108
=
1
[216+(24+3+48)?? ]
 
216+ 75x = 364 
x ~ z 
So the formula would be = C
4
H
6
O
6
 
 
Q4. Mass of sucrose ?? ????
?? ????
?? ????
 produced by mixing ???????? of carbon, ???????? of 
hydrogen and 56 liter ?? ?? at ?? ?????? and ?????? ?? according to given reaction, is 
?? (?? ) + ?? ?? ( ?? ) + ?? ?? ( ?? ) ? ?? ????
?? ????
?? ????
( ?? ) 
(A) 138.5 
(B) 155.5 
(C) 172.5 
(D) 199.5 
Ans: (B) 12C (s) + 11H
2
( g) +
11
2
O
2
( g) ? C
12
H
22
O
11
( s) 
84
12
12
1
56
22.4
7 12 2.5
 
Here O
2
 is limiting reagent 
Moles of C
12
H
22
O
11
 formed =
2.5
11
× 2 =
5
11
 mole 
 Mass =
5
11
× [(12 × 12) + 22 + (11 × 16)] = 155 .45gm 
 
Q5. ?????????? of carbonate of an alkali metal or alkaline earth metal containing some 
inert impurities was made to react with excess ?????? solution. The liberated ????
?? 
occupied 12.315 lit. at ?? ?????? and ?????? ?? . The correct option is 
(A) Mass of impurity is ?????? and metal is Be 
(B) Mass of impurity is ?????? and metal is ???? 
(C) Mass of impurity is ?????? and metal is Be 
(D) Mass of impurity is ?????? and metal is ???? 
Ans: (B) M(CO
3
) ? CO
2
+ MO or 
M
2
(CO
3
) ? CO
2
+ M
2
O 
Mass of CO
3
= 12 + 48 = 60 
Mole of CO
2
=
12.315
(PV)
(RT ) =
12.315
1×(12.315)
× 0.0821× 
300 = 0.5 mole 
Mole of M(CO
3
) or M
2
CO
3
= 0.5 mole 
So, mass of CO
3
 in carbonate = 0.5 × 60 = 30 gram 
Checking all options one by one 
(B) is correct. 
Q7. The mass of ????
?? produced from 620 mixture of ?? ?? ?? ?? ?? ?? and ?? ?? ' prepared 
produce maximum energy is (combustion reaction is exothermic) 
(A) ?????? . ???????? 
(B) ?????? . ???????? 
(C) ?????????? 
(D) ?????????? 
Ans: (C) C
2
H
4
O
2
+ 2O
2
? 2CO
2
+ 2H
2
O 
x gm 620 - ?? gm 
11 11 
?? 60
 mole 
620 - ?? 32
 
 
To produce maximum energy C
2
H
4
O
2
 and O
2
 will be fully consumed. 
?? ?? 60
=
620 - ?? 32
×
1
2
64?? = 37200- 60?? ?? = 300gm
 
Weight of CO
2
= 2 ×
300×44
60
= 440gm 
 
Q8. In the quantitative determination of nitrogen, ?? ?? gas liberated from ?? . ???????? of 
a sample of organic compound was collected over water. If the volume of ?? ?? gas 
collected was 
??????
???? ????
 at total pressure 860 ???????? at ?????? ?? , % by mass of nitrogen in 
the organic compound is 
[Aq. tension at ?????? ?? is ???? ???????? and ?? = ?? . ???? ?? atm ?????? -?? ?? -?? ] 
(A) 
????
?? % 
(B) 
?? ?? % 
(C) 
????
?? % 
(D) 
??????
?? % 
Ans: (A) (Organic compound) +H
2
O ? N
2
 0.42gm 
Moles of N
2
=
PV
RT
=
860
760
×
100
11
×
10
-3
0.08×250
 
=
86
167200
= 5.143× 10
-4
 
Mass of N
2
= 5.143× 10
-4
× 28 = 0.0144gm 
Fraction =
0.0144
0.42
= 0.034 =
10
3
% 
 
Q9. ?????? ???? of ?? . ?????????? and ?????? ???? of ?? . ???? ?? ?? ????
?? are mixed. The normality of the 
resulting mixture is 
(A) ?? . ?? ?? 
(B) ?? . ?? ?? 
(C) ?? . ?? ?? 
(D) ?? . ?? ?? 
Ans: (C) Moles of H
+
= (0.1)(0.3) + (0.2)(0.3) × 2 = 0.15 mole 
Normality =
0.15
500
× 1000 = 0.3 N 
 
Q10. The volume of water which should be added to ?????? ???? of ?? . ???????????? 
solution so as to get a solution of ?? . ???? is 
(A) ?????? ???? 
(B) ?????? ?? ?? 
(C) ?????? ???? 
(D) ?????? ???? 
Ans: (D) Moles of NaOH= (0.300)(0.5) = 0.15 moles 
For molarity = 0.2M =
0.15
 V
 
V = 750 mL 
Volume to be added = 750 - 300 = 450 mL 
 
Q11. The mole fraction of a solution containing ?? . ???????? of urea per ???????????? of 
water would be 
(A) 0.00357 
(B) 0.99643 
(C) 0.00643 
(D) None of these 
Ans: (A) Moles of water =
250
18
= 13.888 mole 
 
 
 
Moles urea =
3
60
= 0.05 mole 
Mole fraction = 0.0036 
 
Q12. The mass of ?? ?? ?? ????
 produced if ?????????? of ?? ?? ?? ?? is mixed with ?????????? of ?? ?? is 
?? ?? ?? ?? + ?? ?? ? ?? ?? ?? ????
+ ????
?? 
(A) ?????????? 
(B) ?????????? 
(C) ?????????? 
(D) ??????????  
Ans: (B) P
4
 S
3
+ 8O
2
? P
4
O
10
+ 3SO
2
 
Moles of O
2
=
384
32
= 12 mole 
Moles of ?? 4
?? 3
=
440
124+96
= 2 mole 
L. R. = O
2
 
So mass of P
4
O
10
 produced 
=
12
8
× [124 + 160] = 426gm 
 
 
Q13. Calculate percentage change in ?? avg 
 of the mixture, if ?????? ?? undergo ???? % 
decomposition. ?????? ?? ? ?????? ?? + ????
?? 
(A) ???? % 
(B) ???? . ???? % 
(C) ???? . ???? % 
(D) Zero