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1 | 905-908 | e , p1
0 < p2
The composition of vapour phase in0 equilibrium with the solution is determined
by the partial pressures of the components If y1 and y2 are the mole fractions of the
Fig |
1 | 906-909 | , p1
0 < p2
The composition of vapour phase in0 equilibrium with the solution is determined
by the partial pressures of the components If y1 and y2 are the mole fractions of the
Fig 1 |
1 | 907-910 | equilibrium with the solution is determined
by the partial pressures of the components If y1 and y2 are the mole fractions of the
Fig 1 3: The plot of vapour pressure and mole
fraction of an ideal solution at constant
temperature |
1 | 908-911 | If y1 and y2 are the mole fractions of the
Fig 1 3: The plot of vapour pressure and mole
fraction of an ideal solution at constant
temperature The dashed lines I and II
represent the partial pressure of the
components |
1 | 909-912 | 1 3: The plot of vapour pressure and mole
fraction of an ideal solution at constant
temperature The dashed lines I and II
represent the partial pressure of the
components (It can be seen from the plot
that p1 and p2 are directly proportional to x1
and x2, respectively) |
1 | 910-913 | 3: The plot of vapour pressure and mole
fraction of an ideal solution at constant
temperature The dashed lines I and II
represent the partial pressure of the
components (It can be seen from the plot
that p1 and p2 are directly proportional to x1
and x2, respectively) The total vapour
pressure is given by line marked III in the
figure |
1 | 911-914 | The dashed lines I and II
represent the partial pressure of the
components (It can be seen from the plot
that p1 and p2 are directly proportional to x1
and x2, respectively) The total vapour
pressure is given by line marked III in the
figure Rationalised 2023-24
11
Solutions
components 1 and 2 respectively in the vapour phase then, using Dalton’s
law of partial pressures:
p1 = y1 ptotal
(1 |
1 | 912-915 | (It can be seen from the plot
that p1 and p2 are directly proportional to x1
and x2, respectively) The total vapour
pressure is given by line marked III in the
figure Rationalised 2023-24
11
Solutions
components 1 and 2 respectively in the vapour phase then, using Dalton’s
law of partial pressures:
p1 = y1 ptotal
(1 17)
p2 = y2 ptotal
(1 |
1 | 913-916 | The total vapour
pressure is given by line marked III in the
figure Rationalised 2023-24
11
Solutions
components 1 and 2 respectively in the vapour phase then, using Dalton’s
law of partial pressures:
p1 = y1 ptotal
(1 17)
p2 = y2 ptotal
(1 18)
In general
pi = yi ptotal
(1 |
1 | 914-917 | Rationalised 2023-24
11
Solutions
components 1 and 2 respectively in the vapour phase then, using Dalton’s
law of partial pressures:
p1 = y1 ptotal
(1 17)
p2 = y2 ptotal
(1 18)
In general
pi = yi ptotal
(1 19)
Vapour pressure of chloroform (CHCl3) and dichloromethane (CH2Cl2)
at 298 K are 200 mm Hg and 415 mm Hg respectively |
1 | 915-918 | 17)
p2 = y2 ptotal
(1 18)
In general
pi = yi ptotal
(1 19)
Vapour pressure of chloroform (CHCl3) and dichloromethane (CH2Cl2)
at 298 K are 200 mm Hg and 415 mm Hg respectively (i) Calculate
the vapour pressure of the solution prepared by mixing 25 |
1 | 916-919 | 18)
In general
pi = yi ptotal
(1 19)
Vapour pressure of chloroform (CHCl3) and dichloromethane (CH2Cl2)
at 298 K are 200 mm Hg and 415 mm Hg respectively (i) Calculate
the vapour pressure of the solution prepared by mixing 25 5 g of
CHCl3 and 40 g of CH2Cl2 at 298 K and, (ii) mole fractions of each
component in vapour phase |
1 | 917-920 | 19)
Vapour pressure of chloroform (CHCl3) and dichloromethane (CH2Cl2)
at 298 K are 200 mm Hg and 415 mm Hg respectively (i) Calculate
the vapour pressure of the solution prepared by mixing 25 5 g of
CHCl3 and 40 g of CH2Cl2 at 298 K and, (ii) mole fractions of each
component in vapour phase (i) Molar mass of CH2Cl2 = 12 × 1 + 1 × 2 + 35 |
1 | 918-921 | (i) Calculate
the vapour pressure of the solution prepared by mixing 25 5 g of
CHCl3 and 40 g of CH2Cl2 at 298 K and, (ii) mole fractions of each
component in vapour phase (i) Molar mass of CH2Cl2 = 12 × 1 + 1 × 2 + 35 5 × 2 = 85 g mol–1
Molar mass of CHCl3
= 12 × 1 + 1 × 1 + 35 |
1 | 919-922 | 5 g of
CHCl3 and 40 g of CH2Cl2 at 298 K and, (ii) mole fractions of each
component in vapour phase (i) Molar mass of CH2Cl2 = 12 × 1 + 1 × 2 + 35 5 × 2 = 85 g mol–1
Molar mass of CHCl3
= 12 × 1 + 1 × 1 + 35 5 × 3 = 119 |
1 | 920-923 | (i) Molar mass of CH2Cl2 = 12 × 1 + 1 × 2 + 35 5 × 2 = 85 g mol–1
Molar mass of CHCl3
= 12 × 1 + 1 × 1 + 35 5 × 3 = 119 5 g mol-1
Moles of CH2Cl2
=
1
40 g
85 g mol = 0 |
1 | 921-924 | 5 × 2 = 85 g mol–1
Molar mass of CHCl3
= 12 × 1 + 1 × 1 + 35 5 × 3 = 119 5 g mol-1
Moles of CH2Cl2
=
1
40 g
85 g mol = 0 47 mol
Moles of CHCl3
=
1
25 |
1 | 922-925 | 5 × 3 = 119 5 g mol-1
Moles of CH2Cl2
=
1
40 g
85 g mol = 0 47 mol
Moles of CHCl3
=
1
25 5 g
119 |
1 | 923-926 | 5 g mol-1
Moles of CH2Cl2
=
1
40 g
85 g mol = 0 47 mol
Moles of CHCl3
=
1
25 5 g
119 5 g mol = 0 |
1 | 924-927 | 47 mol
Moles of CHCl3
=
1
25 5 g
119 5 g mol = 0 213 mol
Total number of moles = 0 |
1 | 925-928 | 5 g
119 5 g mol = 0 213 mol
Total number of moles = 0 47 + 0 |
1 | 926-929 | 5 g mol = 0 213 mol
Total number of moles = 0 47 + 0 213 = 0 |
1 | 927-930 | 213 mol
Total number of moles = 0 47 + 0 213 = 0 683 mol
2
xCH Cl2
= 0 |
1 | 928-931 | 47 + 0 213 = 0 683 mol
2
xCH Cl2
= 0 47 mol
0 |
1 | 929-932 | 213 = 0 683 mol
2
xCH Cl2
= 0 47 mol
0 683 mol
= 0 |
1 | 930-933 | 683 mol
2
xCH Cl2
= 0 47 mol
0 683 mol
= 0 688
xCHCl3
= 1 |
1 | 931-934 | 47 mol
0 683 mol
= 0 688
xCHCl3
= 1 00 – 0 |
1 | 932-935 | 683 mol
= 0 688
xCHCl3
= 1 00 – 0 688 = 0 |
1 | 933-936 | 688
xCHCl3
= 1 00 – 0 688 = 0 312
Using equation (2 |
1 | 934-937 | 00 – 0 688 = 0 312
Using equation (2 16),
ptotal
= p1
0 + (p2
0 – p1
0) x2 = 200 + (415 – 200) × 0 |
1 | 935-938 | 688 = 0 312
Using equation (2 16),
ptotal
= p1
0 + (p2
0 – p1
0) x2 = 200 + (415 – 200) × 0 688
= 200 + 147 |
1 | 936-939 | 312
Using equation (2 16),
ptotal
= p1
0 + (p2
0 – p1
0) x2 = 200 + (415 – 200) × 0 688
= 200 + 147 9 = 347 |
1 | 937-940 | 16),
ptotal
= p1
0 + (p2
0 – p1
0) x2 = 200 + (415 – 200) × 0 688
= 200 + 147 9 = 347 9 mm Hg
(ii) Using the relation (2 |
1 | 938-941 | 688
= 200 + 147 9 = 347 9 mm Hg
(ii) Using the relation (2 19), yi = pi/ptotal, we can calculate the mole
fraction of the components in gas phase (yi) |
1 | 939-942 | 9 = 347 9 mm Hg
(ii) Using the relation (2 19), yi = pi/ptotal, we can calculate the mole
fraction of the components in gas phase (yi) 2
pCH Cl2
= 0 |
1 | 940-943 | 9 mm Hg
(ii) Using the relation (2 19), yi = pi/ptotal, we can calculate the mole
fraction of the components in gas phase (yi) 2
pCH Cl2
= 0 688 × 415 mm Hg = 285 |
1 | 941-944 | 19), yi = pi/ptotal, we can calculate the mole
fraction of the components in gas phase (yi) 2
pCH Cl2
= 0 688 × 415 mm Hg = 285 5 mm Hg
pCHCl3
= 0 |
1 | 942-945 | 2
pCH Cl2
= 0 688 × 415 mm Hg = 285 5 mm Hg
pCHCl3
= 0 312 × 200 mm Hg = 62 |
1 | 943-946 | 688 × 415 mm Hg = 285 5 mm Hg
pCHCl3
= 0 312 × 200 mm Hg = 62 4 mm Hg
2
yCH Cl2
= 285 |
1 | 944-947 | 5 mm Hg
pCHCl3
= 0 312 × 200 mm Hg = 62 4 mm Hg
2
yCH Cl2
= 285 5 mm Hg/347 |
1 | 945-948 | 312 × 200 mm Hg = 62 4 mm Hg
2
yCH Cl2
= 285 5 mm Hg/347 9 mm Hg = 0 |
1 | 946-949 | 4 mm Hg
2
yCH Cl2
= 285 5 mm Hg/347 9 mm Hg = 0 82
yCHCl3
= 62 |
1 | 947-950 | 5 mm Hg/347 9 mm Hg = 0 82
yCHCl3
= 62 4 mm Hg/347 |
1 | 948-951 | 9 mm Hg = 0 82
yCHCl3
= 62 4 mm Hg/347 9 mm Hg = 0 |
1 | 949-952 | 82
yCHCl3
= 62 4 mm Hg/347 9 mm Hg = 0 18
Note: Since, CH2Cl2 is a more volatile component than CHCl3, [
2
p0CH Cl2
=
415 mm Hg and
p0CHCl3
= 200 mm Hg] and the vapour phase is also richer
in CH2Cl2 [
2
yCH Cl2
= 0 |
1 | 950-953 | 4 mm Hg/347 9 mm Hg = 0 18
Note: Since, CH2Cl2 is a more volatile component than CHCl3, [
2
p0CH Cl2
=
415 mm Hg and
p0CHCl3
= 200 mm Hg] and the vapour phase is also richer
in CH2Cl2 [
2
yCH Cl2
= 0 82 and
yCHCl3
= 0 |
1 | 951-954 | 9 mm Hg = 0 18
Note: Since, CH2Cl2 is a more volatile component than CHCl3, [
2
p0CH Cl2
=
415 mm Hg and
p0CHCl3
= 200 mm Hg] and the vapour phase is also richer
in CH2Cl2 [
2
yCH Cl2
= 0 82 and
yCHCl3
= 0 18], it may thus be concluded
that at equilibrium, vapour phase will be always rich in the component
which is more volatile |
1 | 952-955 | 18
Note: Since, CH2Cl2 is a more volatile component than CHCl3, [
2
p0CH Cl2
=
415 mm Hg and
p0CHCl3
= 200 mm Hg] and the vapour phase is also richer
in CH2Cl2 [
2
yCH Cl2
= 0 82 and
yCHCl3
= 0 18], it may thus be concluded
that at equilibrium, vapour phase will be always rich in the component
which is more volatile Example 1 |
1 | 953-956 | 82 and
yCHCl3
= 0 18], it may thus be concluded
that at equilibrium, vapour phase will be always rich in the component
which is more volatile Example 1 5
Example 1 |
1 | 954-957 | 18], it may thus be concluded
that at equilibrium, vapour phase will be always rich in the component
which is more volatile Example 1 5
Example 1 5
Example 1 |
1 | 955-958 | Example 1 5
Example 1 5
Example 1 5
Example 1 |
1 | 956-959 | 5
Example 1 5
Example 1 5
Example 1 5
Example 1 |
1 | 957-960 | 5
Example 1 5
Example 1 5
Example 1 5
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
12
Chemistry
According to Raoult’s law, the vapour pressure of a volatile component
in a given solution is given by pi = xi pi
0 |
1 | 958-961 | 5
Example 1 5
Example 1 5
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
12
Chemistry
According to Raoult’s law, the vapour pressure of a volatile component
in a given solution is given by pi = xi pi
0 In the solution of a gas in a
liquid, one of the components is so volatile that it exists as a gas and
we have already seen that its solubility is given by Henry’s law which
states that
p = KH x |
1 | 959-962 | 5
Example 1 5
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
12
Chemistry
According to Raoult’s law, the vapour pressure of a volatile component
in a given solution is given by pi = xi pi
0 In the solution of a gas in a
liquid, one of the components is so volatile that it exists as a gas and
we have already seen that its solubility is given by Henry’s law which
states that
p = KH x If we compare the equations for Raoult’s law and Henry’s law, it
can be seen that the partial pressure of the volatile component or gas
is directly proportional to its mole fraction in solution |
1 | 960-963 | 5
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
12
Chemistry
According to Raoult’s law, the vapour pressure of a volatile component
in a given solution is given by pi = xi pi
0 In the solution of a gas in a
liquid, one of the components is so volatile that it exists as a gas and
we have already seen that its solubility is given by Henry’s law which
states that
p = KH x If we compare the equations for Raoult’s law and Henry’s law, it
can be seen that the partial pressure of the volatile component or gas
is directly proportional to its mole fraction in solution Only the
proportionality constant KH differs from p1
0 |
1 | 961-964 | In the solution of a gas in a
liquid, one of the components is so volatile that it exists as a gas and
we have already seen that its solubility is given by Henry’s law which
states that
p = KH x If we compare the equations for Raoult’s law and Henry’s law, it
can be seen that the partial pressure of the volatile component or gas
is directly proportional to its mole fraction in solution Only the
proportionality constant KH differs from p1
0 Thus, Raoult’s law becomes
a special case of Henry’s law in which KH becomes equal to p1
0 |
1 | 962-965 | If we compare the equations for Raoult’s law and Henry’s law, it
can be seen that the partial pressure of the volatile component or gas
is directly proportional to its mole fraction in solution Only the
proportionality constant KH differs from p1
0 Thus, Raoult’s law becomes
a special case of Henry’s law in which KH becomes equal to p1
0 Another important class of solutions consists of solids dissolved in
liquid, for example, sodium chloride, glucose, urea and cane sugar in
water and iodine and sulphur dissolved in carbon disulphide |
1 | 963-966 | Only the
proportionality constant KH differs from p1
0 Thus, Raoult’s law becomes
a special case of Henry’s law in which KH becomes equal to p1
0 Another important class of solutions consists of solids dissolved in
liquid, for example, sodium chloride, glucose, urea and cane sugar in
water and iodine and sulphur dissolved in carbon disulphide Some
physical properties of these solutions are quite different from those of
pure solvents |
1 | 964-967 | Thus, Raoult’s law becomes
a special case of Henry’s law in which KH becomes equal to p1
0 Another important class of solutions consists of solids dissolved in
liquid, for example, sodium chloride, glucose, urea and cane sugar in
water and iodine and sulphur dissolved in carbon disulphide Some
physical properties of these solutions are quite different from those of
pure solvents For example, vapour pressure |
1 | 965-968 | Another important class of solutions consists of solids dissolved in
liquid, for example, sodium chloride, glucose, urea and cane sugar in
water and iodine and sulphur dissolved in carbon disulphide Some
physical properties of these solutions are quite different from those of
pure solvents For example, vapour pressure Liquids at a given
temperature
vapourise
and
under
equilibrium conditions the pressure exerted
by the vapours of the liquid over the liquid
phase is called vapour pressure [Fig |
1 | 966-969 | Some
physical properties of these solutions are quite different from those of
pure solvents For example, vapour pressure Liquids at a given
temperature
vapourise
and
under
equilibrium conditions the pressure exerted
by the vapours of the liquid over the liquid
phase is called vapour pressure [Fig 1 |
1 | 967-970 | For example, vapour pressure Liquids at a given
temperature
vapourise
and
under
equilibrium conditions the pressure exerted
by the vapours of the liquid over the liquid
phase is called vapour pressure [Fig 1 4 (a)] |
1 | 968-971 | Liquids at a given
temperature
vapourise
and
under
equilibrium conditions the pressure exerted
by the vapours of the liquid over the liquid
phase is called vapour pressure [Fig 1 4 (a)] In a pure liquid the entire surface is
occupied by the molecules of the liquid |
1 | 969-972 | 1 4 (a)] In a pure liquid the entire surface is
occupied by the molecules of the liquid If a
non-volatile solute is added to a solvent to
give a solution [Fig |
1 | 970-973 | 4 (a)] In a pure liquid the entire surface is
occupied by the molecules of the liquid If a
non-volatile solute is added to a solvent to
give a solution [Fig 1 |
1 | 971-974 | In a pure liquid the entire surface is
occupied by the molecules of the liquid If a
non-volatile solute is added to a solvent to
give a solution [Fig 1 4 |
1 | 972-975 | If a
non-volatile solute is added to a solvent to
give a solution [Fig 1 4 (b)], the vapour
pressure of the solution is solely from the
solvent alone |
1 | 973-976 | 1 4 (b)], the vapour
pressure of the solution is solely from the
solvent alone This vapour pressure of the
solution at a given temperature is found to
be lower than the vapour pressure of the
pure solvent at the same temperature |
1 | 974-977 | 4 (b)], the vapour
pressure of the solution is solely from the
solvent alone This vapour pressure of the
solution at a given temperature is found to
be lower than the vapour pressure of the
pure solvent at the same temperature In
the solution, the surface has both solute and
solvent molecules; thereby the fraction of the
surface covered by the solvent molecules gets
reduced |
1 | 975-978 | (b)], the vapour
pressure of the solution is solely from the
solvent alone This vapour pressure of the
solution at a given temperature is found to
be lower than the vapour pressure of the
pure solvent at the same temperature In
the solution, the surface has both solute and
solvent molecules; thereby the fraction of the
surface covered by the solvent molecules gets
reduced Consequently, the number of
solvent molecules escaping from the surface
is correspondingly reduced, thus, the vapour
pressure is also reduced |
1 | 976-979 | This vapour pressure of the
solution at a given temperature is found to
be lower than the vapour pressure of the
pure solvent at the same temperature In
the solution, the surface has both solute and
solvent molecules; thereby the fraction of the
surface covered by the solvent molecules gets
reduced Consequently, the number of
solvent molecules escaping from the surface
is correspondingly reduced, thus, the vapour
pressure is also reduced The decrease in the vapour pressure of solvent depends on the
quantity of non-volatile solute present in the solution, irrespective of
its nature |
1 | 977-980 | In
the solution, the surface has both solute and
solvent molecules; thereby the fraction of the
surface covered by the solvent molecules gets
reduced Consequently, the number of
solvent molecules escaping from the surface
is correspondingly reduced, thus, the vapour
pressure is also reduced The decrease in the vapour pressure of solvent depends on the
quantity of non-volatile solute present in the solution, irrespective of
its nature For example, decrease in the vapour pressure of water by
adding 1 |
1 | 978-981 | Consequently, the number of
solvent molecules escaping from the surface
is correspondingly reduced, thus, the vapour
pressure is also reduced The decrease in the vapour pressure of solvent depends on the
quantity of non-volatile solute present in the solution, irrespective of
its nature For example, decrease in the vapour pressure of water by
adding 1 0 mol of sucrose to one kg of water is nearly similar to that
produced by adding 1 |
1 | 979-982 | The decrease in the vapour pressure of solvent depends on the
quantity of non-volatile solute present in the solution, irrespective of
its nature For example, decrease in the vapour pressure of water by
adding 1 0 mol of sucrose to one kg of water is nearly similar to that
produced by adding 1 0 mol of urea to the same quantity of water at
the same temperature |
1 | 980-983 | For example, decrease in the vapour pressure of water by
adding 1 0 mol of sucrose to one kg of water is nearly similar to that
produced by adding 1 0 mol of urea to the same quantity of water at
the same temperature Raoult’s law in its general form can be stated as, for any solution
the partial vapour pressure of each volatile component in the
solution is directly proportional to its mole fraction |
1 | 981-984 | 0 mol of sucrose to one kg of water is nearly similar to that
produced by adding 1 0 mol of urea to the same quantity of water at
the same temperature Raoult’s law in its general form can be stated as, for any solution
the partial vapour pressure of each volatile component in the
solution is directly proportional to its mole fraction In a binary solution, let us denote the solvent by 1 and solute by
2 |
1 | 982-985 | 0 mol of urea to the same quantity of water at
the same temperature Raoult’s law in its general form can be stated as, for any solution
the partial vapour pressure of each volatile component in the
solution is directly proportional to its mole fraction In a binary solution, let us denote the solvent by 1 and solute by
2 When the solute is non-volatile, only the solvent molecules are
present in vapour phase and contribute to vapour pressure |
1 | 983-986 | Raoult’s law in its general form can be stated as, for any solution
the partial vapour pressure of each volatile component in the
solution is directly proportional to its mole fraction In a binary solution, let us denote the solvent by 1 and solute by
2 When the solute is non-volatile, only the solvent molecules are
present in vapour phase and contribute to vapour pressure Let p1 be
1 |
1 | 984-987 | In a binary solution, let us denote the solvent by 1 and solute by
2 When the solute is non-volatile, only the solvent molecules are
present in vapour phase and contribute to vapour pressure Let p1 be
1 4 |
1 | 985-988 | When the solute is non-volatile, only the solvent molecules are
present in vapour phase and contribute to vapour pressure Let p1 be
1 4 2 Raoult’s
Law as a
special case
of Henry’s
Law
1 |
1 | 986-989 | Let p1 be
1 4 2 Raoult’s
Law as a
special case
of Henry’s
Law
1 4 |
1 | 987-990 | 4 2 Raoult’s
Law as a
special case
of Henry’s
Law
1 4 3 Vapour
Pressure of
Solutions of
Solids in
Liquids
Fig |
1 | 988-991 | 2 Raoult’s
Law as a
special case
of Henry’s
Law
1 4 3 Vapour
Pressure of
Solutions of
Solids in
Liquids
Fig 1 |
1 | 989-992 | 4 3 Vapour
Pressure of
Solutions of
Solids in
Liquids
Fig 1 4: Decrease in the vapour pressure of the
solvent on account of the presence of
solute in the solvent (a) evaporation of the
molecules of the solvent from its surface
is denoted by
, (b) in a solution, solute
particles have been denoted by and they
also occupy part of the surface area |
1 | 990-993 | 3 Vapour
Pressure of
Solutions of
Solids in
Liquids
Fig 1 4: Decrease in the vapour pressure of the
solvent on account of the presence of
solute in the solvent (a) evaporation of the
molecules of the solvent from its surface
is denoted by
, (b) in a solution, solute
particles have been denoted by and they
also occupy part of the surface area Rationalised 2023-24
13
Solutions
the vapour pressure of the solvent, x1 be
its mole fraction, pi
0 be its vapour pressure
in the pure state |
1 | 991-994 | 1 4: Decrease in the vapour pressure of the
solvent on account of the presence of
solute in the solvent (a) evaporation of the
molecules of the solvent from its surface
is denoted by
, (b) in a solution, solute
particles have been denoted by and they
also occupy part of the surface area Rationalised 2023-24
13
Solutions
the vapour pressure of the solvent, x1 be
its mole fraction, pi
0 be its vapour pressure
in the pure state Then according to
Raoult’s law
p1 µ x1
and
p1 = x1
p10
(1 |
1 | 992-995 | 4: Decrease in the vapour pressure of the
solvent on account of the presence of
solute in the solvent (a) evaporation of the
molecules of the solvent from its surface
is denoted by
, (b) in a solution, solute
particles have been denoted by and they
also occupy part of the surface area Rationalised 2023-24
13
Solutions
the vapour pressure of the solvent, x1 be
its mole fraction, pi
0 be its vapour pressure
in the pure state Then according to
Raoult’s law
p1 µ x1
and
p1 = x1
p10
(1 20)
The proportionality constant is equal
to the vapour pressure of pure solvent,
0
1
p |
1 | 993-996 | Rationalised 2023-24
13
Solutions
the vapour pressure of the solvent, x1 be
its mole fraction, pi
0 be its vapour pressure
in the pure state Then according to
Raoult’s law
p1 µ x1
and
p1 = x1
p10
(1 20)
The proportionality constant is equal
to the vapour pressure of pure solvent,
0
1
p A plot between the vapour pressure and
the mole fraction of the solvent is linear
(Fig |
1 | 994-997 | Then according to
Raoult’s law
p1 µ x1
and
p1 = x1
p10
(1 20)
The proportionality constant is equal
to the vapour pressure of pure solvent,
0
1
p A plot between the vapour pressure and
the mole fraction of the solvent is linear
(Fig 1 |
1 | 995-998 | 20)
The proportionality constant is equal
to the vapour pressure of pure solvent,
0
1
p A plot between the vapour pressure and
the mole fraction of the solvent is linear
(Fig 1 5) |
1 | 996-999 | A plot between the vapour pressure and
the mole fraction of the solvent is linear
(Fig 1 5) Liquid-liquid solutions can be classified into ideal and non-ideal
solutions on the basis of Raoult’s law |
1 | 997-1000 | 1 5) Liquid-liquid solutions can be classified into ideal and non-ideal
solutions on the basis of Raoult’s law The solutions which obey Raoult’s law over the entire range of
concentration are known as ideal solutions |
1 | 998-1001 | 5) Liquid-liquid solutions can be classified into ideal and non-ideal
solutions on the basis of Raoult’s law The solutions which obey Raoult’s law over the entire range of
concentration are known as ideal solutions The ideal solutions have
two other important properties |
1 | 999-1002 | Liquid-liquid solutions can be classified into ideal and non-ideal
solutions on the basis of Raoult’s law The solutions which obey Raoult’s law over the entire range of
concentration are known as ideal solutions The ideal solutions have
two other important properties The enthalpy of mixing of the
pure components to form the solution is zero and the volume
of mixing is also zero, i |
1 | 1000-1003 | The solutions which obey Raoult’s law over the entire range of
concentration are known as ideal solutions The ideal solutions have
two other important properties The enthalpy of mixing of the
pure components to form the solution is zero and the volume
of mixing is also zero, i e |
1 | 1001-1004 | The ideal solutions have
two other important properties The enthalpy of mixing of the
pure components to form the solution is zero and the volume
of mixing is also zero, i e ,
DmixH = 0,
DmixV = 0
(1 |
1 | 1002-1005 | The enthalpy of mixing of the
pure components to form the solution is zero and the volume
of mixing is also zero, i e ,
DmixH = 0,
DmixV = 0
(1 21)
It means that no heat is absorbed or evolved when the components
are mixed |
1 | 1003-1006 | e ,
DmixH = 0,
DmixV = 0
(1 21)
It means that no heat is absorbed or evolved when the components
are mixed Also, the volume of solution would be equal to the sum of
volumes of the two components |
1 | 1004-1007 | ,
DmixH = 0,
DmixV = 0
(1 21)
It means that no heat is absorbed or evolved when the components
are mixed Also, the volume of solution would be equal to the sum of
volumes of the two components At molecular level, ideal behaviour of
the solutions can be explained by considering two components A and
B |
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