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1 | 1205-1208 | 8
DTb = Kb × m = 0 52 K kg mol–1 × 0 1 mol kg–1 = 0 052 K
Since water boils at 373 |
1 | 1206-1209 | 52 K kg mol–1 × 0 1 mol kg–1 = 0 052 K
Since water boils at 373 15 K at 1 |
1 | 1207-1210 | 1 mol kg–1 = 0 052 K
Since water boils at 373 15 K at 1 013 bar pressure, therefore, the
boiling point of solution will be 373 |
1 | 1208-1211 | 052 K
Since water boils at 373 15 K at 1 013 bar pressure, therefore, the
boiling point of solution will be 373 15 + 0 |
1 | 1209-1212 | 15 K at 1 013 bar pressure, therefore, the
boiling point of solution will be 373 15 + 0 052 = 373 |
1 | 1210-1213 | 013 bar pressure, therefore, the
boiling point of solution will be 373 15 + 0 052 = 373 202 K |
1 | 1211-1214 | 15 + 0 052 = 373 202 K The boiling point of benzene is 353 |
1 | 1212-1215 | 052 = 373 202 K The boiling point of benzene is 353 23 K |
1 | 1213-1216 | 202 K The boiling point of benzene is 353 23 K When 1 |
1 | 1214-1217 | The boiling point of benzene is 353 23 K When 1 80 g of a non-volatile
solute was dissolved in 90 g of benzene, the boiling point is raised to
354 |
1 | 1215-1218 | 23 K When 1 80 g of a non-volatile
solute was dissolved in 90 g of benzene, the boiling point is raised to
354 11 K |
1 | 1216-1219 | When 1 80 g of a non-volatile
solute was dissolved in 90 g of benzene, the boiling point is raised to
354 11 K Calculate the molar mass of the solute |
1 | 1217-1220 | 80 g of a non-volatile
solute was dissolved in 90 g of benzene, the boiling point is raised to
354 11 K Calculate the molar mass of the solute Kb for benzene is 2 |
1 | 1218-1221 | 11 K Calculate the molar mass of the solute Kb for benzene is 2 53
K kg mol–1
The elevation (DTb) in the boiling point = 354 |
1 | 1219-1222 | Calculate the molar mass of the solute Kb for benzene is 2 53
K kg mol–1
The elevation (DTb) in the boiling point = 354 11 K – 353 |
1 | 1220-1223 | Kb for benzene is 2 53
K kg mol–1
The elevation (DTb) in the boiling point = 354 11 K – 353 23 K = 0 |
1 | 1221-1224 | 53
K kg mol–1
The elevation (DTb) in the boiling point = 354 11 K – 353 23 K = 0 88 K
Substituting these values in expression (2 |
1 | 1222-1225 | 11 K – 353 23 K = 0 88 K
Substituting these values in expression (2 33) we get
M2 =
–1
–1
2 |
1 | 1223-1226 | 23 K = 0 88 K
Substituting these values in expression (2 33) we get
M2 =
–1
–1
2 53 K kg mol
× 1 |
1 | 1224-1227 | 88 K
Substituting these values in expression (2 33) we get
M2 =
–1
–1
2 53 K kg mol
× 1 8 g × 1000 g kg
0 |
1 | 1225-1228 | 33) we get
M2 =
–1
–1
2 53 K kg mol
× 1 8 g × 1000 g kg
0 88 K × 90 g
= 58 g mol–1
Therefore, molar mass of the solute, M2 = 58 g mol–1
Fig |
1 | 1226-1229 | 53 K kg mol
× 1 8 g × 1000 g kg
0 88 K × 90 g
= 58 g mol–1
Therefore, molar mass of the solute, M2 = 58 g mol–1
Fig 1 |
1 | 1227-1230 | 8 g × 1000 g kg
0 88 K × 90 g
= 58 g mol–1
Therefore, molar mass of the solute, M2 = 58 g mol–1
Fig 1 8: Diagram showing DTf, depression
of the freezing point of a solvent in
a solution |
1 | 1228-1231 | 88 K × 90 g
= 58 g mol–1
Therefore, molar mass of the solute, M2 = 58 g mol–1
Fig 1 8: Diagram showing DTf, depression
of the freezing point of a solvent in
a solution Solution
Solution
Solution
Solution
Solution
1 |
1 | 1229-1232 | 1 8: Diagram showing DTf, depression
of the freezing point of a solvent in
a solution Solution
Solution
Solution
Solution
Solution
1 6 |
1 | 1230-1233 | 8: Diagram showing DTf, depression
of the freezing point of a solvent in
a solution Solution
Solution
Solution
Solution
Solution
1 6 3 Depression
of Freezing
Point
Rationalised 2023-24
19
Solutions
Depression Constant or Cryoscopic Constant |
1 | 1231-1234 | Solution
Solution
Solution
Solution
Solution
1 6 3 Depression
of Freezing
Point
Rationalised 2023-24
19
Solutions
Depression Constant or Cryoscopic Constant The unit of Kf is K kg
mol-1 |
1 | 1232-1235 | 6 3 Depression
of Freezing
Point
Rationalised 2023-24
19
Solutions
Depression Constant or Cryoscopic Constant The unit of Kf is K kg
mol-1 Values of Kf for some common solvents are listed in Table 1 |
1 | 1233-1236 | 3 Depression
of Freezing
Point
Rationalised 2023-24
19
Solutions
Depression Constant or Cryoscopic Constant The unit of Kf is K kg
mol-1 Values of Kf for some common solvents are listed in Table 1 3 |
1 | 1234-1237 | The unit of Kf is K kg
mol-1 Values of Kf for some common solvents are listed in Table 1 3 If w2 gram of the solute having molar mass as M2, present in w1
gram of solvent, produces the depression in freezing point DTf of the
solvent then molality of the solute is given by the equation (1 |
1 | 1235-1238 | Values of Kf for some common solvents are listed in Table 1 3 If w2 gram of the solute having molar mass as M2, present in w1
gram of solvent, produces the depression in freezing point DTf of the
solvent then molality of the solute is given by the equation (1 31) |
1 | 1236-1239 | 3 If w2 gram of the solute having molar mass as M2, present in w1
gram of solvent, produces the depression in freezing point DTf of the
solvent then molality of the solute is given by the equation (1 31) m
= w
w
2
2
1
/
/1000
M
(1 |
1 | 1237-1240 | If w2 gram of the solute having molar mass as M2, present in w1
gram of solvent, produces the depression in freezing point DTf of the
solvent then molality of the solute is given by the equation (1 31) m
= w
w
2
2
1
/
/1000
M
(1 31)
Substituting this value of molality in equation (1 |
1 | 1238-1241 | 31) m
= w
w
2
2
1
/
/1000
M
(1 31)
Substituting this value of molality in equation (1 34) we get:
DTf =
f
2
2
1
/
/1000
K
M
w
w
DTf =
f
2
2
1
×
× 1000
×
K
M
w
w
(1 |
1 | 1239-1242 | m
= w
w
2
2
1
/
/1000
M
(1 31)
Substituting this value of molality in equation (1 34) we get:
DTf =
f
2
2
1
/
/1000
K
M
w
w
DTf =
f
2
2
1
×
× 1000
×
K
M
w
w
(1 35)
M2 =
f
2
f
1
×
× 1000
×
K
T
w
w
(1 |
1 | 1240-1243 | 31)
Substituting this value of molality in equation (1 34) we get:
DTf =
f
2
2
1
/
/1000
K
M
w
w
DTf =
f
2
2
1
×
× 1000
×
K
M
w
w
(1 35)
M2 =
f
2
f
1
×
× 1000
×
K
T
w
w
(1 36)
Thus for determining the molar mass of the solute we should know
the quantities w1, w2, DTf, along with the molal freezing point depression
constant |
1 | 1241-1244 | 34) we get:
DTf =
f
2
2
1
/
/1000
K
M
w
w
DTf =
f
2
2
1
×
× 1000
×
K
M
w
w
(1 35)
M2 =
f
2
f
1
×
× 1000
×
K
T
w
w
(1 36)
Thus for determining the molar mass of the solute we should know
the quantities w1, w2, DTf, along with the molal freezing point depression
constant The values of Kf and Kb, which depend upon the nature of the
solvent, can be ascertained from the following relations |
1 | 1242-1245 | 35)
M2 =
f
2
f
1
×
× 1000
×
K
T
w
w
(1 36)
Thus for determining the molar mass of the solute we should know
the quantities w1, w2, DTf, along with the molal freezing point depression
constant The values of Kf and Kb, which depend upon the nature of the
solvent, can be ascertained from the following relations Kf
=
2
1
f
fus
×
×
1000 ×
R
M
T
H
(1 |
1 | 1243-1246 | 36)
Thus for determining the molar mass of the solute we should know
the quantities w1, w2, DTf, along with the molal freezing point depression
constant The values of Kf and Kb, which depend upon the nature of the
solvent, can be ascertained from the following relations Kf
=
2
1
f
fus
×
×
1000 ×
R
M
T
H
(1 37)
Kb
=
2
1
b
vap
×
1000 × ×
R
M
T
H
(1 |
1 | 1244-1247 | The values of Kf and Kb, which depend upon the nature of the
solvent, can be ascertained from the following relations Kf
=
2
1
f
fus
×
×
1000 ×
R
M
T
H
(1 37)
Kb
=
2
1
b
vap
×
1000 × ×
R
M
T
H
(1 38)
Here the symbols R and M1 stand for the gas constant and molar
mass of the solvent, respectively and Tf and Tb denote the freezing point
and the boiling point of the pure solvent respectively in kelvin |
1 | 1245-1248 | Kf
=
2
1
f
fus
×
×
1000 ×
R
M
T
H
(1 37)
Kb
=
2
1
b
vap
×
1000 × ×
R
M
T
H
(1 38)
Here the symbols R and M1 stand for the gas constant and molar
mass of the solvent, respectively and Tf and Tb denote the freezing point
and the boiling point of the pure solvent respectively in kelvin Further,
DfusH and DvapH represent the enthalpies for the fusion and vapourisation
of the solvent, respectively |
1 | 1246-1249 | 37)
Kb
=
2
1
b
vap
×
1000 × ×
R
M
T
H
(1 38)
Here the symbols R and M1 stand for the gas constant and molar
mass of the solvent, respectively and Tf and Tb denote the freezing point
and the boiling point of the pure solvent respectively in kelvin Further,
DfusH and DvapH represent the enthalpies for the fusion and vapourisation
of the solvent, respectively Solvent
b |
1 | 1247-1250 | 38)
Here the symbols R and M1 stand for the gas constant and molar
mass of the solvent, respectively and Tf and Tb denote the freezing point
and the boiling point of the pure solvent respectively in kelvin Further,
DfusH and DvapH represent the enthalpies for the fusion and vapourisation
of the solvent, respectively Solvent
b p |
1 | 1248-1251 | Further,
DfusH and DvapH represent the enthalpies for the fusion and vapourisation
of the solvent, respectively Solvent
b p /K
Kb/K kg mol-1
f |
1 | 1249-1252 | Solvent
b p /K
Kb/K kg mol-1
f p |
1 | 1250-1253 | p /K
Kb/K kg mol-1
f p /K
Kf/K kg mol-1
Water
373 |
1 | 1251-1254 | /K
Kb/K kg mol-1
f p /K
Kf/K kg mol-1
Water
373 15
0 |
1 | 1252-1255 | p /K
Kf/K kg mol-1
Water
373 15
0 52
273 |
1 | 1253-1256 | /K
Kf/K kg mol-1
Water
373 15
0 52
273 0
1 |
1 | 1254-1257 | 15
0 52
273 0
1 86
Ethanol
351 |
1 | 1255-1258 | 52
273 0
1 86
Ethanol
351 5
1 |
1 | 1256-1259 | 0
1 86
Ethanol
351 5
1 20
155 |
1 | 1257-1260 | 86
Ethanol
351 5
1 20
155 7
1 |
1 | 1258-1261 | 5
1 20
155 7
1 99
Cyclohexane
353 |
1 | 1259-1262 | 20
155 7
1 99
Cyclohexane
353 74
2 |
1 | 1260-1263 | 7
1 99
Cyclohexane
353 74
2 79
279 |
1 | 1261-1264 | 99
Cyclohexane
353 74
2 79
279 55
20 |
1 | 1262-1265 | 74
2 79
279 55
20 00
Benzene
353 |
1 | 1263-1266 | 79
279 55
20 00
Benzene
353 3
2 |
1 | 1264-1267 | 55
20 00
Benzene
353 3
2 53
278 |
1 | 1265-1268 | 00
Benzene
353 3
2 53
278 6
5 |
1 | 1266-1269 | 3
2 53
278 6
5 12
Chloroform
334 |
1 | 1267-1270 | 53
278 6
5 12
Chloroform
334 4
3 |
1 | 1268-1271 | 6
5 12
Chloroform
334 4
3 63
209 |
1 | 1269-1272 | 12
Chloroform
334 4
3 63
209 6
4 |
1 | 1270-1273 | 4
3 63
209 6
4 79
Carbon tetrachloride
350 |
1 | 1271-1274 | 63
209 6
4 79
Carbon tetrachloride
350 0
5 |
1 | 1272-1275 | 6
4 79
Carbon tetrachloride
350 0
5 03
250 |
1 | 1273-1276 | 79
Carbon tetrachloride
350 0
5 03
250 5
31 |
1 | 1274-1277 | 0
5 03
250 5
31 8
Carbon disulphide
319 |
1 | 1275-1278 | 03
250 5
31 8
Carbon disulphide
319 4
2 |
1 | 1276-1279 | 5
31 8
Carbon disulphide
319 4
2 34
164 |
1 | 1277-1280 | 8
Carbon disulphide
319 4
2 34
164 2
3 |
1 | 1278-1281 | 4
2 34
164 2
3 83
Diethyl ether
307 |
1 | 1279-1282 | 34
164 2
3 83
Diethyl ether
307 8
2 |
1 | 1280-1283 | 2
3 83
Diethyl ether
307 8
2 02
156 |
1 | 1281-1284 | 83
Diethyl ether
307 8
2 02
156 9
1 |
1 | 1282-1285 | 8
2 02
156 9
1 79
Acetic acid
391 |
1 | 1283-1286 | 02
156 9
1 79
Acetic acid
391 1
2 |
1 | 1284-1287 | 9
1 79
Acetic acid
391 1
2 93
290 |
1 | 1285-1288 | 79
Acetic acid
391 1
2 93
290 0
3 |
1 | 1286-1289 | 1
2 93
290 0
3 90
Table 1 |
1 | 1287-1290 | 93
290 0
3 90
Table 1 3: Molal Boiling Point Elevation and Freezing Point
Depression Constants for Some Solvents
Rationalised 2023-24
20
Chemistry
Fig |
1 | 1288-1291 | 0
3 90
Table 1 3: Molal Boiling Point Elevation and Freezing Point
Depression Constants for Some Solvents
Rationalised 2023-24
20
Chemistry
Fig 1 |
1 | 1289-1292 | 90
Table 1 3: Molal Boiling Point Elevation and Freezing Point
Depression Constants for Some Solvents
Rationalised 2023-24
20
Chemistry
Fig 1 9
Level of solution
rises in the thistle
funnel due to
osmosis of solvent |
1 | 1290-1293 | 3: Molal Boiling Point Elevation and Freezing Point
Depression Constants for Some Solvents
Rationalised 2023-24
20
Chemistry
Fig 1 9
Level of solution
rises in the thistle
funnel due to
osmosis of solvent 45 g of ethylene glycol (C2H6O2) is mixed with 600 g of water |
1 | 1291-1294 | 1 9
Level of solution
rises in the thistle
funnel due to
osmosis of solvent 45 g of ethylene glycol (C2H6O2) is mixed with 600 g of water Calculate
(a) the freezing point depression and (b) the freezing point of the solution |
1 | 1292-1295 | 9
Level of solution
rises in the thistle
funnel due to
osmosis of solvent 45 g of ethylene glycol (C2H6O2) is mixed with 600 g of water Calculate
(a) the freezing point depression and (b) the freezing point of the solution Depression in freezing point is related to the molality, therefore, the molality
of the solution with respect to ethylene glycol =
moles of ethylene glycol
mass of water in kilogram
Moles of ethylene glycol =
1
45 g
62 g mol = 0 |
1 | 1293-1296 | 45 g of ethylene glycol (C2H6O2) is mixed with 600 g of water Calculate
(a) the freezing point depression and (b) the freezing point of the solution Depression in freezing point is related to the molality, therefore, the molality
of the solution with respect to ethylene glycol =
moles of ethylene glycol
mass of water in kilogram
Moles of ethylene glycol =
1
45 g
62 g mol = 0 73 mol
Mass of water in kg =
1
600g
1000g kg = 0 |
1 | 1294-1297 | Calculate
(a) the freezing point depression and (b) the freezing point of the solution Depression in freezing point is related to the molality, therefore, the molality
of the solution with respect to ethylene glycol =
moles of ethylene glycol
mass of water in kilogram
Moles of ethylene glycol =
1
45 g
62 g mol = 0 73 mol
Mass of water in kg =
1
600g
1000g kg = 0 6 kg
Hence molality of ethylene glycol =
0 |
1 | 1295-1298 | Depression in freezing point is related to the molality, therefore, the molality
of the solution with respect to ethylene glycol =
moles of ethylene glycol
mass of water in kilogram
Moles of ethylene glycol =
1
45 g
62 g mol = 0 73 mol
Mass of water in kg =
1
600g
1000g kg = 0 6 kg
Hence molality of ethylene glycol =
0 73 mol
0 |
1 | 1296-1299 | 73 mol
Mass of water in kg =
1
600g
1000g kg = 0 6 kg
Hence molality of ethylene glycol =
0 73 mol
0 60 kg = 1 |
1 | 1297-1300 | 6 kg
Hence molality of ethylene glycol =
0 73 mol
0 60 kg = 1 2 mol kg –1
Therefore freezing point depression,
ÄTf = 1 |
1 | 1298-1301 | 73 mol
0 60 kg = 1 2 mol kg –1
Therefore freezing point depression,
ÄTf = 1 86 K kg mol–1 × 1 |
1 | 1299-1302 | 60 kg = 1 2 mol kg –1
Therefore freezing point depression,
ÄTf = 1 86 K kg mol–1 × 1 2 mol kg –1 = 2 |
1 | 1300-1303 | 2 mol kg –1
Therefore freezing point depression,
ÄTf = 1 86 K kg mol–1 × 1 2 mol kg –1 = 2 2 K
Freezing point of the aqueous solution = 273 |
1 | 1301-1304 | 86 K kg mol–1 × 1 2 mol kg –1 = 2 2 K
Freezing point of the aqueous solution = 273 15 K – 2 |
1 | 1302-1305 | 2 mol kg –1 = 2 2 K
Freezing point of the aqueous solution = 273 15 K – 2 2 K = 270 |
1 | 1303-1306 | 2 K
Freezing point of the aqueous solution = 273 15 K – 2 2 K = 270 95 K
1 |
1 | 1304-1307 | 15 K – 2 2 K = 270 95 K
1 00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the
freezing point of benzene by 0 |
Subsets and Splits