Chapter
stringclasses 18
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stringlengths 3
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stringlengths 7
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|---|---|---|
1 | 1505-1508 | 98
2 00
MgSO4
1 21
1 53
1 |
1 | 1506-1509 | 00
MgSO4
1 21
1 53
1 82
2 |
1 | 1507-1510 | 21
1 53
1 82
2 00
K2SO4
2 |
1 | 1508-1511 | 53
1 82
2 00
K2SO4
2 32
2 |
1 | 1509-1512 | 82
2 00
K2SO4
2 32
2 70
2 |
1 | 1510-1513 | 00
K2SO4
2 32
2 70
2 84
3 |
1 | 1511-1514 | 32
2 70
2 84
3 00
* represent i values for incomplete dissociation |
1 | 1512-1515 | 70
2 84
3 00
* represent i values for incomplete dissociation Table 1 |
1 | 1513-1516 | 84
3 00
* represent i values for incomplete dissociation Table 1 4: Values of van’t Hoff factor, i, at Various Concentrations
for NaCl, KCl, MgSO4 and K2SO4 |
1 | 1514-1517 | 00
* represent i values for incomplete dissociation Table 1 4: Values of van’t Hoff factor, i, at Various Concentrations
for NaCl, KCl, MgSO4 and K2SO4 Rationalised 2023-24
26
Chemistry
Example 1 |
1 | 1515-1518 | Table 1 4: Values of van’t Hoff factor, i, at Various Concentrations
for NaCl, KCl, MgSO4 and K2SO4 Rationalised 2023-24
26
Chemistry
Example 1 13
Example 1 |
1 | 1516-1519 | 4: Values of van’t Hoff factor, i, at Various Concentrations
for NaCl, KCl, MgSO4 and K2SO4 Rationalised 2023-24
26
Chemistry
Example 1 13
Example 1 13
Example 1 |
1 | 1517-1520 | Rationalised 2023-24
26
Chemistry
Example 1 13
Example 1 13
Example 1 13
Example 1 |
1 | 1518-1521 | 13
Example 1 13
Example 1 13
Example 1 13
Example 1 |
1 | 1519-1522 | 13
Example 1 13
Example 1 13
Example 1 13
=
1
1
122 g mol
241 |
1 | 1520-1523 | 13
Example 1 13
Example 1 13
=
1
1
122 g mol
241 98 g mol
or
2
x =
122
1
1
0 |
1 | 1521-1524 | 13
Example 1 13
=
1
1
122 g mol
241 98 g mol
or
2
x =
122
1
1
0 504
0 |
1 | 1522-1525 | 13
=
1
1
122 g mol
241 98 g mol
or
2
x =
122
1
1
0 504
0 496
241 |
1 | 1523-1526 | 98 g mol
or
2
x =
122
1
1
0 504
0 496
241 98
or
x
= 2 × 0 |
1 | 1524-1527 | 504
0 496
241 98
or
x
= 2 × 0 496 = 0 |
1 | 1525-1528 | 496
241 98
or
x
= 2 × 0 496 = 0 992
Therefore, degree of association of benzoic acid in benzene is 99 |
1 | 1526-1529 | 98
or
x
= 2 × 0 496 = 0 992
Therefore, degree of association of benzoic acid in benzene is 99 2 % |
1 | 1527-1530 | 496 = 0 992
Therefore, degree of association of benzoic acid in benzene is 99 2 % 0 |
1 | 1528-1531 | 992
Therefore, degree of association of benzoic acid in benzene is 99 2 % 0 6 mL of acetic acid (CH3COOH), having density 1 |
1 | 1529-1532 | 2 % 0 6 mL of acetic acid (CH3COOH), having density 1 06 g mL–1, is
dissolved in 1 litre of water |
1 | 1530-1533 | 0 6 mL of acetic acid (CH3COOH), having density 1 06 g mL–1, is
dissolved in 1 litre of water The depression in freezing point
observed for this strength of acid was 0 |
1 | 1531-1534 | 6 mL of acetic acid (CH3COOH), having density 1 06 g mL–1, is
dissolved in 1 litre of water The depression in freezing point
observed for this strength of acid was 0 0205°C |
1 | 1532-1535 | 06 g mL–1, is
dissolved in 1 litre of water The depression in freezing point
observed for this strength of acid was 0 0205°C Calculate the van’t
Hoff factor and the dissociation constant of acid |
1 | 1533-1536 | The depression in freezing point
observed for this strength of acid was 0 0205°C Calculate the van’t
Hoff factor and the dissociation constant of acid Number of moles of acetic acid
=
1
1
0 |
1 | 1534-1537 | 0205°C Calculate the van’t
Hoff factor and the dissociation constant of acid Number of moles of acetic acid
=
1
1
0 6 mL
1 |
1 | 1535-1538 | Calculate the van’t
Hoff factor and the dissociation constant of acid Number of moles of acetic acid
=
1
1
0 6 mL
1 06 g mL
60 g mol
= 0 |
1 | 1536-1539 | Number of moles of acetic acid
=
1
1
0 6 mL
1 06 g mL
60 g mol
= 0 0106 mol = n
Molality =
1
0 |
1 | 1537-1540 | 6 mL
1 06 g mL
60 g mol
= 0 0106 mol = n
Molality =
1
0 0106 mol
1000 mL
1 g mL
= 0 |
1 | 1538-1541 | 06 g mL
60 g mol
= 0 0106 mol = n
Molality =
1
0 0106 mol
1000 mL
1 g mL
= 0 0106 mol kg–1
Using equation (1 |
1 | 1539-1542 | 0106 mol = n
Molality =
1
0 0106 mol
1000 mL
1 g mL
= 0 0106 mol kg–1
Using equation (1 35)
DTf = 1 |
1 | 1540-1543 | 0106 mol
1000 mL
1 g mL
= 0 0106 mol kg–1
Using equation (1 35)
DTf = 1 86 K kg mol–1 × 0 |
1 | 1541-1544 | 0106 mol kg–1
Using equation (1 35)
DTf = 1 86 K kg mol–1 × 0 0106 mol kg–1 = 0 |
1 | 1542-1545 | 35)
DTf = 1 86 K kg mol–1 × 0 0106 mol kg–1 = 0 0197 K
van’t Hoff Factor (i) =
Observed freezing point
Calculated freezing point = 0 |
1 | 1543-1546 | 86 K kg mol–1 × 0 0106 mol kg–1 = 0 0197 K
van’t Hoff Factor (i) =
Observed freezing point
Calculated freezing point = 0 0205 K
0 |
1 | 1544-1547 | 0106 mol kg–1 = 0 0197 K
van’t Hoff Factor (i) =
Observed freezing point
Calculated freezing point = 0 0205 K
0 0197 K = 1 |
1 | 1545-1548 | 0197 K
van’t Hoff Factor (i) =
Observed freezing point
Calculated freezing point = 0 0205 K
0 0197 K = 1 041
Acetic acid is a weak electrolyte and will dissociate into two ions:
acetate and hydrogen ions per molecule of acetic acid |
1 | 1546-1549 | 0205 K
0 0197 K = 1 041
Acetic acid is a weak electrolyte and will dissociate into two ions:
acetate and hydrogen ions per molecule of acetic acid If x is the
degree of dissociation of acetic acid, then we would have n (1 – x)
moles of undissociated acetic acid, nx moles of CH3COO– and nx
moles of H+ ions,
(
)
+
−
+
−
⇌
3
3
CH COOH
H
CH COO
mol
0
0
mol
mol
nn1
nx
nx
x
Thus total moles of particles are: n(1 – x + x + x) = n(1 + x )
1
1
1 |
1 | 1547-1550 | 0197 K = 1 041
Acetic acid is a weak electrolyte and will dissociate into two ions:
acetate and hydrogen ions per molecule of acetic acid If x is the
degree of dissociation of acetic acid, then we would have n (1 – x)
moles of undissociated acetic acid, nx moles of CH3COO– and nx
moles of H+ ions,
(
)
+
−
+
−
⇌
3
3
CH COOH
H
CH COO
mol
0
0
mol
mol
nn1
nx
nx
x
Thus total moles of particles are: n(1 – x + x + x) = n(1 + x )
1
1
1 041
n
x
i
x
n
Thus degree of dissociation of acetic acid = x = 1 |
1 | 1548-1551 | 041
Acetic acid is a weak electrolyte and will dissociate into two ions:
acetate and hydrogen ions per molecule of acetic acid If x is the
degree of dissociation of acetic acid, then we would have n (1 – x)
moles of undissociated acetic acid, nx moles of CH3COO– and nx
moles of H+ ions,
(
)
+
−
+
−
⇌
3
3
CH COOH
H
CH COO
mol
0
0
mol
mol
nn1
nx
nx
x
Thus total moles of particles are: n(1 – x + x + x) = n(1 + x )
1
1
1 041
n
x
i
x
n
Thus degree of dissociation of acetic acid = x = 1 041– 1 |
1 | 1549-1552 | If x is the
degree of dissociation of acetic acid, then we would have n (1 – x)
moles of undissociated acetic acid, nx moles of CH3COO– and nx
moles of H+ ions,
(
)
+
−
+
−
⇌
3
3
CH COOH
H
CH COO
mol
0
0
mol
mol
nn1
nx
nx
x
Thus total moles of particles are: n(1 – x + x + x) = n(1 + x )
1
1
1 041
n
x
i
x
n
Thus degree of dissociation of acetic acid = x = 1 041– 1 000 = 0 |
1 | 1550-1553 | 041
n
x
i
x
n
Thus degree of dissociation of acetic acid = x = 1 041– 1 000 = 0 041
Then
[CH3COOH] = n(1 – x) = 0 |
1 | 1551-1554 | 041– 1 000 = 0 041
Then
[CH3COOH] = n(1 – x) = 0 0106 (1 – 0 |
1 | 1552-1555 | 000 = 0 041
Then
[CH3COOH] = n(1 – x) = 0 0106 (1 – 0 041),
[CH3COO–] = nx = 0 |
1 | 1553-1556 | 041
Then
[CH3COOH] = n(1 – x) = 0 0106 (1 – 0 041),
[CH3COO–] = nx = 0 0106 × 0 |
1 | 1554-1557 | 0106 (1 – 0 041),
[CH3COO–] = nx = 0 0106 × 0 041, [H+] = nx = 0 |
1 | 1555-1558 | 041),
[CH3COO–] = nx = 0 0106 × 0 041, [H+] = nx = 0 0106 × 0 |
1 | 1556-1559 | 0106 × 0 041, [H+] = nx = 0 0106 × 0 041 |
1 | 1557-1560 | 041, [H+] = nx = 0 0106 × 0 041 Ka =
3
3
[
][
]
[
]
CH COO
H
CH COOH
=
0 |
1 | 1558-1561 | 0106 × 0 041 Ka =
3
3
[
][
]
[
]
CH COO
H
CH COOH
=
0 0106 × 0 |
1 | 1559-1562 | 041 Ka =
3
3
[
][
]
[
]
CH COO
H
CH COOH
=
0 0106 × 0 041 × 0 |
1 | 1560-1563 | Ka =
3
3
[
][
]
[
]
CH COO
H
CH COOH
=
0 0106 × 0 041 × 0 0106 × 0 |
1 | 1561-1564 | 0106 × 0 041 × 0 0106 × 0 041
0 |
1 | 1562-1565 | 041 × 0 0106 × 0 041
0 0106 (1 |
1 | 1563-1566 | 0106 × 0 041
0 0106 (1 00
0 |
1 | 1564-1567 | 041
0 0106 (1 00
0 041)
= 1 |
1 | 1565-1568 | 0106 (1 00
0 041)
= 1 86 × 10–5
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
27
Solutions
Summary
Summary
Summary
Summary
Summary
A solution is a homogeneous mixture of two or more substances |
1 | 1566-1569 | 00
0 041)
= 1 86 × 10–5
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
27
Solutions
Summary
Summary
Summary
Summary
Summary
A solution is a homogeneous mixture of two or more substances Solutions are
classified as solid, liquid and gaseous solutions |
1 | 1567-1570 | 041)
= 1 86 × 10–5
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
27
Solutions
Summary
Summary
Summary
Summary
Summary
A solution is a homogeneous mixture of two or more substances Solutions are
classified as solid, liquid and gaseous solutions The concentration of a solution is
expressed in terms of mole fraction, molarity, molality and in percentages |
1 | 1568-1571 | 86 × 10–5
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
27
Solutions
Summary
Summary
Summary
Summary
Summary
A solution is a homogeneous mixture of two or more substances Solutions are
classified as solid, liquid and gaseous solutions The concentration of a solution is
expressed in terms of mole fraction, molarity, molality and in percentages The
dissolution of a gas in a liquid is governed by Henry’s law, according to which, at a
given temperature, the solubility of a gas in a liquid is directly proportional to
the partial pressure of the gas |
1 | 1569-1572 | Solutions are
classified as solid, liquid and gaseous solutions The concentration of a solution is
expressed in terms of mole fraction, molarity, molality and in percentages The
dissolution of a gas in a liquid is governed by Henry’s law, according to which, at a
given temperature, the solubility of a gas in a liquid is directly proportional to
the partial pressure of the gas The vapour pressure of the solvent is lowered by
the presence of a non-volatile solute in the solution and this lowering of vapour
pressure of the solvent is governed by Raoult’s law, according to which the relative
lowering of vapour pressure of the solvent over a solution is equal to the mole
fraction of a non-volatile solute present in the solution |
1 | 1570-1573 | The concentration of a solution is
expressed in terms of mole fraction, molarity, molality and in percentages The
dissolution of a gas in a liquid is governed by Henry’s law, according to which, at a
given temperature, the solubility of a gas in a liquid is directly proportional to
the partial pressure of the gas The vapour pressure of the solvent is lowered by
the presence of a non-volatile solute in the solution and this lowering of vapour
pressure of the solvent is governed by Raoult’s law, according to which the relative
lowering of vapour pressure of the solvent over a solution is equal to the mole
fraction of a non-volatile solute present in the solution However, in a binary
liquid solution, if both the components of the solution are volatile then another
form of Raoult’s law is used |
1 | 1571-1574 | The
dissolution of a gas in a liquid is governed by Henry’s law, according to which, at a
given temperature, the solubility of a gas in a liquid is directly proportional to
the partial pressure of the gas The vapour pressure of the solvent is lowered by
the presence of a non-volatile solute in the solution and this lowering of vapour
pressure of the solvent is governed by Raoult’s law, according to which the relative
lowering of vapour pressure of the solvent over a solution is equal to the mole
fraction of a non-volatile solute present in the solution However, in a binary
liquid solution, if both the components of the solution are volatile then another
form of Raoult’s law is used Mathematically, this form of the Raoult’s law is stated
as:
0
0
total
1
2
2
1
p
p x
p x |
1 | 1572-1575 | The vapour pressure of the solvent is lowered by
the presence of a non-volatile solute in the solution and this lowering of vapour
pressure of the solvent is governed by Raoult’s law, according to which the relative
lowering of vapour pressure of the solvent over a solution is equal to the mole
fraction of a non-volatile solute present in the solution However, in a binary
liquid solution, if both the components of the solution are volatile then another
form of Raoult’s law is used Mathematically, this form of the Raoult’s law is stated
as:
0
0
total
1
2
2
1
p
p x
p x Solutions which obey Raoult’s law over the entire range
of concentration are called ideal solutions |
1 | 1573-1576 | However, in a binary
liquid solution, if both the components of the solution are volatile then another
form of Raoult’s law is used Mathematically, this form of the Raoult’s law is stated
as:
0
0
total
1
2
2
1
p
p x
p x Solutions which obey Raoult’s law over the entire range
of concentration are called ideal solutions Two types of deviations from Raoult’s
law, called positive and negative deviations are observed |
1 | 1574-1577 | Mathematically, this form of the Raoult’s law is stated
as:
0
0
total
1
2
2
1
p
p x
p x Solutions which obey Raoult’s law over the entire range
of concentration are called ideal solutions Two types of deviations from Raoult’s
law, called positive and negative deviations are observed Azeotropes arise due to
very large deviations from Raoult’s law |
1 | 1575-1578 | Solutions which obey Raoult’s law over the entire range
of concentration are called ideal solutions Two types of deviations from Raoult’s
law, called positive and negative deviations are observed Azeotropes arise due to
very large deviations from Raoult’s law The properties of solutions which depend on the number of solute particles and
are independent of their chemical identity are called colligative properties |
1 | 1576-1579 | Two types of deviations from Raoult’s
law, called positive and negative deviations are observed Azeotropes arise due to
very large deviations from Raoult’s law The properties of solutions which depend on the number of solute particles and
are independent of their chemical identity are called colligative properties These
are lowering of vapour pressure, elevation of boiling point, depression of freezing
point and osmotic pressure |
1 | 1577-1580 | Azeotropes arise due to
very large deviations from Raoult’s law The properties of solutions which depend on the number of solute particles and
are independent of their chemical identity are called colligative properties These
are lowering of vapour pressure, elevation of boiling point, depression of freezing
point and osmotic pressure The process of osmosis can be reversed if a pressure
higher than the osmotic pressure is applied to the solution |
1 | 1578-1581 | The properties of solutions which depend on the number of solute particles and
are independent of their chemical identity are called colligative properties These
are lowering of vapour pressure, elevation of boiling point, depression of freezing
point and osmotic pressure The process of osmosis can be reversed if a pressure
higher than the osmotic pressure is applied to the solution Colligative properties
have been used to determine the molar mass of solutes |
1 | 1579-1582 | These
are lowering of vapour pressure, elevation of boiling point, depression of freezing
point and osmotic pressure The process of osmosis can be reversed if a pressure
higher than the osmotic pressure is applied to the solution Colligative properties
have been used to determine the molar mass of solutes Solutes which dissociate in
solution exhibit molar mass lower than the actual molar mass and those which
associate show higher molar mass than their actual values |
1 | 1580-1583 | The process of osmosis can be reversed if a pressure
higher than the osmotic pressure is applied to the solution Colligative properties
have been used to determine the molar mass of solutes Solutes which dissociate in
solution exhibit molar mass lower than the actual molar mass and those which
associate show higher molar mass than their actual values Quantitatively, the extent to which a solute is dissociated or associated can be
expressed by van’t Hoff factor i |
1 | 1581-1584 | Colligative properties
have been used to determine the molar mass of solutes Solutes which dissociate in
solution exhibit molar mass lower than the actual molar mass and those which
associate show higher molar mass than their actual values Quantitatively, the extent to which a solute is dissociated or associated can be
expressed by van’t Hoff factor i This factor has been defined as ratio of normal
molar mass to experimentally determined molar mass or as the ratio of observed
colligative property to the calculated colligative property |
1 | 1582-1585 | Solutes which dissociate in
solution exhibit molar mass lower than the actual molar mass and those which
associate show higher molar mass than their actual values Quantitatively, the extent to which a solute is dissociated or associated can be
expressed by van’t Hoff factor i This factor has been defined as ratio of normal
molar mass to experimentally determined molar mass or as the ratio of observed
colligative property to the calculated colligative property 1 |
1 | 1583-1586 | Quantitatively, the extent to which a solute is dissociated or associated can be
expressed by van’t Hoff factor i This factor has been defined as ratio of normal
molar mass to experimentally determined molar mass or as the ratio of observed
colligative property to the calculated colligative property 1 1
Define the term solution |
1 | 1584-1587 | This factor has been defined as ratio of normal
molar mass to experimentally determined molar mass or as the ratio of observed
colligative property to the calculated colligative property 1 1
Define the term solution How many types of solutions are formed |
1 | 1585-1588 | 1 1
Define the term solution How many types of solutions are formed Write briefly
about each type with an example |
1 | 1586-1589 | 1
Define the term solution How many types of solutions are formed Write briefly
about each type with an example 1 |
1 | 1587-1590 | How many types of solutions are formed Write briefly
about each type with an example 1 2
Give an example of a solid solution in which the solute is a gas |
1 | 1588-1591 | Write briefly
about each type with an example 1 2
Give an example of a solid solution in which the solute is a gas 1 |
1 | 1589-1592 | 1 2
Give an example of a solid solution in which the solute is a gas 1 3
Define the following terms:
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage |
1 | 1590-1593 | 2
Give an example of a solid solution in which the solute is a gas 1 3
Define the following terms:
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage 1 |
1 | 1591-1594 | 1 3
Define the following terms:
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage 1 4
Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in
aqueous solution |
1 | 1592-1595 | 3
Define the following terms:
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage 1 4
Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in
aqueous solution What should be the molarity of such a sample of the acid if
the density of the solution is 1 |
1 | 1593-1596 | 1 4
Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in
aqueous solution What should be the molarity of such a sample of the acid if
the density of the solution is 1 504 g mL–1 |
1 | 1594-1597 | 4
Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in
aqueous solution What should be the molarity of such a sample of the acid if
the density of the solution is 1 504 g mL–1 Exercises
Exercises
Exercises
Exercises
Exercises
Rationalised 2023-24
28
Chemistry
1 |
1 | 1595-1598 | What should be the molarity of such a sample of the acid if
the density of the solution is 1 504 g mL–1 Exercises
Exercises
Exercises
Exercises
Exercises
Rationalised 2023-24
28
Chemistry
1 5
A solution of glucose in water is labelled as 10% w/w, what would be the
molality and mole fraction of each component in the solution |
1 | 1596-1599 | 504 g mL–1 Exercises
Exercises
Exercises
Exercises
Exercises
Rationalised 2023-24
28
Chemistry
1 5
A solution of glucose in water is labelled as 10% w/w, what would be the
molality and mole fraction of each component in the solution If the density of
solution is 1 |
1 | 1597-1600 | Exercises
Exercises
Exercises
Exercises
Exercises
Rationalised 2023-24
28
Chemistry
1 5
A solution of glucose in water is labelled as 10% w/w, what would be the
molality and mole fraction of each component in the solution If the density of
solution is 1 2 g mL–1, then what shall be the molarity of the solution |
1 | 1598-1601 | 5
A solution of glucose in water is labelled as 10% w/w, what would be the
molality and mole fraction of each component in the solution If the density of
solution is 1 2 g mL–1, then what shall be the molarity of the solution 1 |
1 | 1599-1602 | If the density of
solution is 1 2 g mL–1, then what shall be the molarity of the solution 1 6
How many mL of 0 |
1 | 1600-1603 | 2 g mL–1, then what shall be the molarity of the solution 1 6
How many mL of 0 1 M HCl are required to react completely with 1 g mixture
of Na2CO3 and NaHCO3 containing equimolar amounts of both |
1 | 1601-1604 | 1 6
How many mL of 0 1 M HCl are required to react completely with 1 g mixture
of Na2CO3 and NaHCO3 containing equimolar amounts of both 1 |
1 | 1602-1605 | 6
How many mL of 0 1 M HCl are required to react completely with 1 g mixture
of Na2CO3 and NaHCO3 containing equimolar amounts of both 1 7
A solution is obtained by mixing 300 g of 25% solution and 400 g of 40%
solution by mass |
1 | 1603-1606 | 1 M HCl are required to react completely with 1 g mixture
of Na2CO3 and NaHCO3 containing equimolar amounts of both 1 7
A solution is obtained by mixing 300 g of 25% solution and 400 g of 40%
solution by mass Calculate the mass percentage of the resulting solution |
1 | 1604-1607 | 1 7
A solution is obtained by mixing 300 g of 25% solution and 400 g of 40%
solution by mass Calculate the mass percentage of the resulting solution 1 |
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