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1 | 1105-1108 | 26)
Here n1 and n2 are the number of moles of solvent and solute
respectively present in the solution For dilute solutions n2 < < n1,
hence neglecting n2 in the denominator we have
10
1
0
1
p
p
p
=
2
1
n
n
(1 27)
or
10
1
0
1
-
p
p
p
=
2
1
2
1
w ×
× w
M
M
(1 28)
Here w1 and w2 are the masses and M1 and M2 are the molar masses
of the solvent and solute respectively |
1 | 1106-1109 | For dilute solutions n2 < < n1,
hence neglecting n2 in the denominator we have
10
1
0
1
p
p
p
=
2
1
n
n
(1 27)
or
10
1
0
1
-
p
p
p
=
2
1
2
1
w ×
× w
M
M
(1 28)
Here w1 and w2 are the masses and M1 and M2 are the molar masses
of the solvent and solute respectively From this equation (1 |
1 | 1107-1110 | 27)
or
10
1
0
1
-
p
p
p
=
2
1
2
1
w ×
× w
M
M
(1 28)
Here w1 and w2 are the masses and M1 and M2 are the molar masses
of the solvent and solute respectively From this equation (1 28), knowing all other quantities, the molar
mass of solute (M2) can be calculated |
1 | 1108-1111 | 28)
Here w1 and w2 are the masses and M1 and M2 are the molar masses
of the solvent and solute respectively From this equation (1 28), knowing all other quantities, the molar
mass of solute (M2) can be calculated Example 1 |
1 | 1109-1112 | From this equation (1 28), knowing all other quantities, the molar
mass of solute (M2) can be calculated Example 1 6
Example 1 |
1 | 1110-1113 | 28), knowing all other quantities, the molar
mass of solute (M2) can be calculated Example 1 6
Example 1 6
Example 1 |
1 | 1111-1114 | Example 1 6
Example 1 6
Example 1 6
Example 1 |
1 | 1112-1115 | 6
Example 1 6
Example 1 6
Example 1 6
Example 1 |
1 | 1113-1116 | 6
Example 1 6
Example 1 6
Example 1 6
The vapour pressure of pure benzene at a certain temperature is 0 |
1 | 1114-1117 | 6
Example 1 6
Example 1 6
The vapour pressure of pure benzene at a certain temperature is 0 850
bar |
1 | 1115-1118 | 6
Example 1 6
The vapour pressure of pure benzene at a certain temperature is 0 850
bar A non-volatile, non-electrolyte solid weighing 0 |
1 | 1116-1119 | 6
The vapour pressure of pure benzene at a certain temperature is 0 850
bar A non-volatile, non-electrolyte solid weighing 0 5 g when added to
39 |
1 | 1117-1120 | 850
bar A non-volatile, non-electrolyte solid weighing 0 5 g when added to
39 0 g of benzene (molar mass 78 g mol-1) |
1 | 1118-1121 | A non-volatile, non-electrolyte solid weighing 0 5 g when added to
39 0 g of benzene (molar mass 78 g mol-1) Vapour pressure of the solution,
then, is 0 |
1 | 1119-1122 | 5 g when added to
39 0 g of benzene (molar mass 78 g mol-1) Vapour pressure of the solution,
then, is 0 845 bar |
1 | 1120-1123 | 0 g of benzene (molar mass 78 g mol-1) Vapour pressure of the solution,
then, is 0 845 bar What is the molar mass of the solid substance |
1 | 1121-1124 | Vapour pressure of the solution,
then, is 0 845 bar What is the molar mass of the solid substance The various quantities known to us are as follows:
p1
0 = 0 |
1 | 1122-1125 | 845 bar What is the molar mass of the solid substance The various quantities known to us are as follows:
p1
0 = 0 850 bar; p = 0 |
1 | 1123-1126 | What is the molar mass of the solid substance The various quantities known to us are as follows:
p1
0 = 0 850 bar; p = 0 845 bar; M1 = 78 g mol–1; w2 = 0 |
1 | 1124-1127 | The various quantities known to us are as follows:
p1
0 = 0 850 bar; p = 0 845 bar; M1 = 78 g mol–1; w2 = 0 5 g; w1 = 39 g
Substituting these values in equation (2 |
1 | 1125-1128 | 850 bar; p = 0 845 bar; M1 = 78 g mol–1; w2 = 0 5 g; w1 = 39 g
Substituting these values in equation (2 28), we get
0 |
1 | 1126-1129 | 845 bar; M1 = 78 g mol–1; w2 = 0 5 g; w1 = 39 g
Substituting these values in equation (2 28), we get
0 850 bar – 0 |
1 | 1127-1130 | 5 g; w1 = 39 g
Substituting these values in equation (2 28), we get
0 850 bar – 0 845 bar
0 |
1 | 1128-1131 | 28), we get
0 850 bar – 0 845 bar
0 850 bar
=
–1
2
0 |
1 | 1129-1132 | 850 bar – 0 845 bar
0 850 bar
=
–1
2
0 5 g × 78 g mol
× 39 g
M
Therefore, M2 = 170 g mol–1
The vapour pressure of a liquid increases with increase of
temperature |
1 | 1130-1133 | 845 bar
0 850 bar
=
–1
2
0 5 g × 78 g mol
× 39 g
M
Therefore, M2 = 170 g mol–1
The vapour pressure of a liquid increases with increase of
temperature It boils at the temperature at which its vapour pressure
is equal to the atmospheric pressure |
1 | 1131-1134 | 850 bar
=
–1
2
0 5 g × 78 g mol
× 39 g
M
Therefore, M2 = 170 g mol–1
The vapour pressure of a liquid increases with increase of
temperature It boils at the temperature at which its vapour pressure
is equal to the atmospheric pressure For example, water boils at
373 |
1 | 1132-1135 | 5 g × 78 g mol
× 39 g
M
Therefore, M2 = 170 g mol–1
The vapour pressure of a liquid increases with increase of
temperature It boils at the temperature at which its vapour pressure
is equal to the atmospheric pressure For example, water boils at
373 15 K (100° C) because at this temperature the vapour pressure
of water is 1 |
1 | 1133-1136 | It boils at the temperature at which its vapour pressure
is equal to the atmospheric pressure For example, water boils at
373 15 K (100° C) because at this temperature the vapour pressure
of water is 1 013 bar (1 atmosphere) |
1 | 1134-1137 | For example, water boils at
373 15 K (100° C) because at this temperature the vapour pressure
of water is 1 013 bar (1 atmosphere) We have also learnt in the last
section that vapour pressure of the solvent decreases in the presence
of non-volatile solute |
1 | 1135-1138 | 15 K (100° C) because at this temperature the vapour pressure
of water is 1 013 bar (1 atmosphere) We have also learnt in the last
section that vapour pressure of the solvent decreases in the presence
of non-volatile solute Fig |
1 | 1136-1139 | 013 bar (1 atmosphere) We have also learnt in the last
section that vapour pressure of the solvent decreases in the presence
of non-volatile solute Fig 1 |
1 | 1137-1140 | We have also learnt in the last
section that vapour pressure of the solvent decreases in the presence
of non-volatile solute Fig 1 7 depicts the variation of vapour pressure
of the pure solvent and solution as a function of temperature |
1 | 1138-1141 | Fig 1 7 depicts the variation of vapour pressure
of the pure solvent and solution as a function of temperature For
example, the vapour pressure of an aqueous solution of sucrose is
less than 1 |
1 | 1139-1142 | 1 7 depicts the variation of vapour pressure
of the pure solvent and solution as a function of temperature For
example, the vapour pressure of an aqueous solution of sucrose is
less than 1 013 bar at 373 |
1 | 1140-1143 | 7 depicts the variation of vapour pressure
of the pure solvent and solution as a function of temperature For
example, the vapour pressure of an aqueous solution of sucrose is
less than 1 013 bar at 373 15 K |
1 | 1141-1144 | For
example, the vapour pressure of an aqueous solution of sucrose is
less than 1 013 bar at 373 15 K In order to make this solution
boil, its vapour pressure must be increased to 1 |
1 | 1142-1145 | 013 bar at 373 15 K In order to make this solution
boil, its vapour pressure must be increased to 1 013 bar by
raising the temperature above the boiling temperature of
the pure solvent (water) |
1 | 1143-1146 | 15 K In order to make this solution
boil, its vapour pressure must be increased to 1 013 bar by
raising the temperature above the boiling temperature of
the pure solvent (water) Thus, the boiling point of a solution is
1 |
1 | 1144-1147 | In order to make this solution
boil, its vapour pressure must be increased to 1 013 bar by
raising the temperature above the boiling temperature of
the pure solvent (water) Thus, the boiling point of a solution is
1 6 |
1 | 1145-1148 | 013 bar by
raising the temperature above the boiling temperature of
the pure solvent (water) Thus, the boiling point of a solution is
1 6 2 Elevation of
Boiling Point
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
17
Solutions
always higher than that of the boiling point of
the pure solvent in which the solution is prepared
as shown in Fig |
1 | 1146-1149 | Thus, the boiling point of a solution is
1 6 2 Elevation of
Boiling Point
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
17
Solutions
always higher than that of the boiling point of
the pure solvent in which the solution is prepared
as shown in Fig 1 |
1 | 1147-1150 | 6 2 Elevation of
Boiling Point
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
17
Solutions
always higher than that of the boiling point of
the pure solvent in which the solution is prepared
as shown in Fig 1 7 |
1 | 1148-1151 | 2 Elevation of
Boiling Point
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
17
Solutions
always higher than that of the boiling point of
the pure solvent in which the solution is prepared
as shown in Fig 1 7 Similar to lowering of vapour
pressure, the elevation of boiling point also
depends on the number of solute molecules
rather than their nature |
1 | 1149-1152 | 1 7 Similar to lowering of vapour
pressure, the elevation of boiling point also
depends on the number of solute molecules
rather than their nature A solution of 1 mol of
sucrose in 1000 g of water boils at 373 |
1 | 1150-1153 | 7 Similar to lowering of vapour
pressure, the elevation of boiling point also
depends on the number of solute molecules
rather than their nature A solution of 1 mol of
sucrose in 1000 g of water boils at 373 52 K at
one atmospheric pressure |
1 | 1151-1154 | Similar to lowering of vapour
pressure, the elevation of boiling point also
depends on the number of solute molecules
rather than their nature A solution of 1 mol of
sucrose in 1000 g of water boils at 373 52 K at
one atmospheric pressure Let
b0
T be the boiling point of pure solvent and
b
T be the boiling point of solution |
1 | 1152-1155 | A solution of 1 mol of
sucrose in 1000 g of water boils at 373 52 K at
one atmospheric pressure Let
b0
T be the boiling point of pure solvent and
b
T be the boiling point of solution The increase in
the boiling point
0
b
b
b
T
T
T is known as
elevation of boiling point |
1 | 1153-1156 | 52 K at
one atmospheric pressure Let
b0
T be the boiling point of pure solvent and
b
T be the boiling point of solution The increase in
the boiling point
0
b
b
b
T
T
T is known as
elevation of boiling point Experiments have shown that for dilute
solutions the elevation of boiling point (DTb) is
directly proportional to the molal concentration of
the solute in a solution |
1 | 1154-1157 | Let
b0
T be the boiling point of pure solvent and
b
T be the boiling point of solution The increase in
the boiling point
0
b
b
b
T
T
T is known as
elevation of boiling point Experiments have shown that for dilute
solutions the elevation of boiling point (DTb) is
directly proportional to the molal concentration of
the solute in a solution Thus
DTb µ m
(1 |
1 | 1155-1158 | The increase in
the boiling point
0
b
b
b
T
T
T is known as
elevation of boiling point Experiments have shown that for dilute
solutions the elevation of boiling point (DTb) is
directly proportional to the molal concentration of
the solute in a solution Thus
DTb µ m
(1 29)
or
DTb = Kb m
(1 |
1 | 1156-1159 | Experiments have shown that for dilute
solutions the elevation of boiling point (DTb) is
directly proportional to the molal concentration of
the solute in a solution Thus
DTb µ m
(1 29)
or
DTb = Kb m
(1 30)
Here m (molality) is the number of moles of solute dissolved in 1 kg
of solvent and the constant of proportionality, Kb is called Boiling Point
Elevation Constant or Molal Elevation Constant (Ebullioscopic
Constant) |
1 | 1157-1160 | Thus
DTb µ m
(1 29)
or
DTb = Kb m
(1 30)
Here m (molality) is the number of moles of solute dissolved in 1 kg
of solvent and the constant of proportionality, Kb is called Boiling Point
Elevation Constant or Molal Elevation Constant (Ebullioscopic
Constant) The unit of Kb is K kg mol-1 |
1 | 1158-1161 | 29)
or
DTb = Kb m
(1 30)
Here m (molality) is the number of moles of solute dissolved in 1 kg
of solvent and the constant of proportionality, Kb is called Boiling Point
Elevation Constant or Molal Elevation Constant (Ebullioscopic
Constant) The unit of Kb is K kg mol-1 Values of Kb for some common
solvents are given in Table 1 |
1 | 1159-1162 | 30)
Here m (molality) is the number of moles of solute dissolved in 1 kg
of solvent and the constant of proportionality, Kb is called Boiling Point
Elevation Constant or Molal Elevation Constant (Ebullioscopic
Constant) The unit of Kb is K kg mol-1 Values of Kb for some common
solvents are given in Table 1 3 |
1 | 1160-1163 | The unit of Kb is K kg mol-1 Values of Kb for some common
solvents are given in Table 1 3 If w2 gram of solute of molar mass M2
is dissolved in w1 gram of solvent, then molality, m of the solution is
given by the expression:
m
=
2
2
1
/M
/1000
ww
=
2
2
1
1000 ×
×
M
ww
(1 |
1 | 1161-1164 | Values of Kb for some common
solvents are given in Table 1 3 If w2 gram of solute of molar mass M2
is dissolved in w1 gram of solvent, then molality, m of the solution is
given by the expression:
m
=
2
2
1
/M
/1000
ww
=
2
2
1
1000 ×
×
M
ww
(1 31)
Substituting the value of molality in equation (1 |
1 | 1162-1165 | 3 If w2 gram of solute of molar mass M2
is dissolved in w1 gram of solvent, then molality, m of the solution is
given by the expression:
m
=
2
2
1
/M
/1000
ww
=
2
2
1
1000 ×
×
M
ww
(1 31)
Substituting the value of molality in equation (1 30) we get
DTb =
b
2
2
1
× 1000 ×
×
K
M
w
w
(1 |
1 | 1163-1166 | If w2 gram of solute of molar mass M2
is dissolved in w1 gram of solvent, then molality, m of the solution is
given by the expression:
m
=
2
2
1
/M
/1000
ww
=
2
2
1
1000 ×
×
M
ww
(1 31)
Substituting the value of molality in equation (1 30) we get
DTb =
b
2
2
1
× 1000 ×
×
K
M
w
w
(1 32)
M2 =
2
b
b
1
1000 ×
×
×
K
T
w
w
(1 |
1 | 1164-1167 | 31)
Substituting the value of molality in equation (1 30) we get
DTb =
b
2
2
1
× 1000 ×
×
K
M
w
w
(1 32)
M2 =
2
b
b
1
1000 ×
×
×
K
T
w
w
(1 33)
Thus, in order to determine M2, molar mass of the solute, known
mass of solute in a known mass of the solvent is taken and DTb is
determined experimentally for a known solvent whose Kb value is known |
1 | 1165-1168 | 30) we get
DTb =
b
2
2
1
× 1000 ×
×
K
M
w
w
(1 32)
M2 =
2
b
b
1
1000 ×
×
×
K
T
w
w
(1 33)
Thus, in order to determine M2, molar mass of the solute, known
mass of solute in a known mass of the solvent is taken and DTb is
determined experimentally for a known solvent whose Kb value is known 18 g of glucose, C6H12O6, is dissolved in 1 kg of water in a saucepan |
1 | 1166-1169 | 32)
M2 =
2
b
b
1
1000 ×
×
×
K
T
w
w
(1 33)
Thus, in order to determine M2, molar mass of the solute, known
mass of solute in a known mass of the solvent is taken and DTb is
determined experimentally for a known solvent whose Kb value is known 18 g of glucose, C6H12O6, is dissolved in 1 kg of water in a saucepan At what temperature will water boil at 1 |
1 | 1167-1170 | 33)
Thus, in order to determine M2, molar mass of the solute, known
mass of solute in a known mass of the solvent is taken and DTb is
determined experimentally for a known solvent whose Kb value is known 18 g of glucose, C6H12O6, is dissolved in 1 kg of water in a saucepan At what temperature will water boil at 1 013 bar |
1 | 1168-1171 | 18 g of glucose, C6H12O6, is dissolved in 1 kg of water in a saucepan At what temperature will water boil at 1 013 bar Kb for water is 0 |
1 | 1169-1172 | At what temperature will water boil at 1 013 bar Kb for water is 0 52
K kg mol-1 |
1 | 1170-1173 | 013 bar Kb for water is 0 52
K kg mol-1 Moles of glucose = 18 g/ 180 g mol–1 = 0 |
1 | 1171-1174 | Kb for water is 0 52
K kg mol-1 Moles of glucose = 18 g/ 180 g mol–1 = 0 1 mol
Number of kilograms of solvent = 1 kg
Thus molality of glucose solution = 0 |
1 | 1172-1175 | 52
K kg mol-1 Moles of glucose = 18 g/ 180 g mol–1 = 0 1 mol
Number of kilograms of solvent = 1 kg
Thus molality of glucose solution = 0 1 mol kg-1
For water, change in boiling point
Example 1 |
1 | 1173-1176 | Moles of glucose = 18 g/ 180 g mol–1 = 0 1 mol
Number of kilograms of solvent = 1 kg
Thus molality of glucose solution = 0 1 mol kg-1
For water, change in boiling point
Example 1 7
Example 1 |
1 | 1174-1177 | 1 mol
Number of kilograms of solvent = 1 kg
Thus molality of glucose solution = 0 1 mol kg-1
For water, change in boiling point
Example 1 7
Example 1 7
Example 1 |
1 | 1175-1178 | 1 mol kg-1
For water, change in boiling point
Example 1 7
Example 1 7
Example 1 7
Example 1 |
1 | 1176-1179 | 7
Example 1 7
Example 1 7
Example 1 7
Example 1 |
1 | 1177-1180 | 7
Example 1 7
Example 1 7
Example 1 7
Solution
Solution
Solution
Solution
Solution
Fig |
1 | 1178-1181 | 7
Example 1 7
Example 1 7
Solution
Solution
Solution
Solution
Solution
Fig 1 |
1 | 1179-1182 | 7
Example 1 7
Solution
Solution
Solution
Solution
Solution
Fig 1 7: The vapour pressure curve for
solution lies below the curve for pure
water |
1 | 1180-1183 | 7
Solution
Solution
Solution
Solution
Solution
Fig 1 7: The vapour pressure curve for
solution lies below the curve for pure
water The diagram shows that DTb
denotes the elevation of boiling
point of a solvent in solution |
1 | 1181-1184 | 1 7: The vapour pressure curve for
solution lies below the curve for pure
water The diagram shows that DTb
denotes the elevation of boiling
point of a solvent in solution 1 |
1 | 1182-1185 | 7: The vapour pressure curve for
solution lies below the curve for pure
water The diagram shows that DTb
denotes the elevation of boiling
point of a solvent in solution 1 013 bar
or
Rationalised 2023-24
18
Chemistry
The lowering of vapour pressure of a solution causes a lowering of the
freezing point compared to that of the pure solvent (Fig |
1 | 1183-1186 | The diagram shows that DTb
denotes the elevation of boiling
point of a solvent in solution 1 013 bar
or
Rationalised 2023-24
18
Chemistry
The lowering of vapour pressure of a solution causes a lowering of the
freezing point compared to that of the pure solvent (Fig 1 |
1 | 1184-1187 | 1 013 bar
or
Rationalised 2023-24
18
Chemistry
The lowering of vapour pressure of a solution causes a lowering of the
freezing point compared to that of the pure solvent (Fig 1 8) |
1 | 1185-1188 | 013 bar
or
Rationalised 2023-24
18
Chemistry
The lowering of vapour pressure of a solution causes a lowering of the
freezing point compared to that of the pure solvent (Fig 1 8) We know
that at the freezing point of a substance, the solid phase is in dynamic
equilibrium with the liquid phase |
1 | 1186-1189 | 1 8) We know
that at the freezing point of a substance, the solid phase is in dynamic
equilibrium with the liquid phase Thus, the
freezing point of a substance may be defined as
the temperature at which the vapour pressure of
the substance in its liquid phase is equal to its
vapour pressure in the solid phase |
1 | 1187-1190 | 8) We know
that at the freezing point of a substance, the solid phase is in dynamic
equilibrium with the liquid phase Thus, the
freezing point of a substance may be defined as
the temperature at which the vapour pressure of
the substance in its liquid phase is equal to its
vapour pressure in the solid phase A solution
will freeze when its vapour pressure equals the
vapour pressure of the pure solid solvent as is
clear from Fig |
1 | 1188-1191 | We know
that at the freezing point of a substance, the solid phase is in dynamic
equilibrium with the liquid phase Thus, the
freezing point of a substance may be defined as
the temperature at which the vapour pressure of
the substance in its liquid phase is equal to its
vapour pressure in the solid phase A solution
will freeze when its vapour pressure equals the
vapour pressure of the pure solid solvent as is
clear from Fig 1 |
1 | 1189-1192 | Thus, the
freezing point of a substance may be defined as
the temperature at which the vapour pressure of
the substance in its liquid phase is equal to its
vapour pressure in the solid phase A solution
will freeze when its vapour pressure equals the
vapour pressure of the pure solid solvent as is
clear from Fig 1 8 |
1 | 1190-1193 | A solution
will freeze when its vapour pressure equals the
vapour pressure of the pure solid solvent as is
clear from Fig 1 8 According to Raoult’s law,
when a non-volatile solid is added to the solvent
its vapour pressure decreases and now it would
become equal to that of solid solvent at lower
temperature |
1 | 1191-1194 | 1 8 According to Raoult’s law,
when a non-volatile solid is added to the solvent
its vapour pressure decreases and now it would
become equal to that of solid solvent at lower
temperature Thus, the freezing point of the
solvent decreases |
1 | 1192-1195 | 8 According to Raoult’s law,
when a non-volatile solid is added to the solvent
its vapour pressure decreases and now it would
become equal to that of solid solvent at lower
temperature Thus, the freezing point of the
solvent decreases Let
0
fT be the freezing point of pure solvent
and
fT be its freezing point when non-volatile
solute is dissolved in it |
1 | 1193-1196 | According to Raoult’s law,
when a non-volatile solid is added to the solvent
its vapour pressure decreases and now it would
become equal to that of solid solvent at lower
temperature Thus, the freezing point of the
solvent decreases Let
0
fT be the freezing point of pure solvent
and
fT be its freezing point when non-volatile
solute is dissolved in it The decrease in freezing
point |
1 | 1194-1197 | Thus, the freezing point of the
solvent decreases Let
0
fT be the freezing point of pure solvent
and
fT be its freezing point when non-volatile
solute is dissolved in it The decrease in freezing
point 0
f
f
f
T
T
T is known as depression in
freezing point |
1 | 1195-1198 | Let
0
fT be the freezing point of pure solvent
and
fT be its freezing point when non-volatile
solute is dissolved in it The decrease in freezing
point 0
f
f
f
T
T
T is known as depression in
freezing point Similar to elevation of boiling point, depression of freezing point (DTf)
for dilute solution (ideal solution) is directly proportional to molality,
m of the solution |
1 | 1196-1199 | The decrease in freezing
point 0
f
f
f
T
T
T is known as depression in
freezing point Similar to elevation of boiling point, depression of freezing point (DTf)
for dilute solution (ideal solution) is directly proportional to molality,
m of the solution Thus,
DTf µ m
or
DTf = Kf m
(1 |
1 | 1197-1200 | 0
f
f
f
T
T
T is known as depression in
freezing point Similar to elevation of boiling point, depression of freezing point (DTf)
for dilute solution (ideal solution) is directly proportional to molality,
m of the solution Thus,
DTf µ m
or
DTf = Kf m
(1 34)
The proportionality constant, Kf, which depends on the nature of the
solvent is known as Freezing Point Depression Constant or Molal
Example 1 |
1 | 1198-1201 | Similar to elevation of boiling point, depression of freezing point (DTf)
for dilute solution (ideal solution) is directly proportional to molality,
m of the solution Thus,
DTf µ m
or
DTf = Kf m
(1 34)
The proportionality constant, Kf, which depends on the nature of the
solvent is known as Freezing Point Depression Constant or Molal
Example 1 8
Example 1 |
1 | 1199-1202 | Thus,
DTf µ m
or
DTf = Kf m
(1 34)
The proportionality constant, Kf, which depends on the nature of the
solvent is known as Freezing Point Depression Constant or Molal
Example 1 8
Example 1 8
Example 1 |
1 | 1200-1203 | 34)
The proportionality constant, Kf, which depends on the nature of the
solvent is known as Freezing Point Depression Constant or Molal
Example 1 8
Example 1 8
Example 1 8
Example 1 |
1 | 1201-1204 | 8
Example 1 8
Example 1 8
Example 1 8
Example 1 |
1 | 1202-1205 | 8
Example 1 8
Example 1 8
Example 1 8
DTb = Kb × m = 0 |
1 | 1203-1206 | 8
Example 1 8
Example 1 8
DTb = Kb × m = 0 52 K kg mol–1 × 0 |
1 | 1204-1207 | 8
Example 1 8
DTb = Kb × m = 0 52 K kg mol–1 × 0 1 mol kg–1 = 0 |
Subsets and Splits