Chapter
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1
|
1605-1608
|
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 8
An antifreeze solution is prepared from 222
|
1
|
1606-1609
|
Calculate the mass percentage of the resulting solution 1 8
An antifreeze solution is prepared from 222 6 g of ethylene glycol (C2H6O2) and
200 g of water
|
1
|
1607-1610
|
1 8
An antifreeze solution is prepared from 222 6 g of ethylene glycol (C2H6O2) and
200 g of water Calculate the molality of the solution
|
1
|
1608-1611
|
8
An antifreeze solution is prepared from 222 6 g of ethylene glycol (C2H6O2) and
200 g of water Calculate the molality of the solution If the density of the
solution is 1
|
1
|
1609-1612
|
6 g of ethylene glycol (C2H6O2) and
200 g of water Calculate the molality of the solution If the density of the
solution is 1 072 g mL–1, then what shall be the molarity of the solution
|
1
|
1610-1613
|
Calculate the molality of the solution If the density of the
solution is 1 072 g mL–1, then what shall be the molarity of the solution 1
|
1
|
1611-1614
|
If the density of the
solution is 1 072 g mL–1, then what shall be the molarity of the solution 1 9
A sample of drinking water was found to be severely contaminated with
chloroform (CHCl3) supposed to be a carcinogen
|
1
|
1612-1615
|
072 g mL–1, then what shall be the molarity of the solution 1 9
A sample of drinking water was found to be severely contaminated with
chloroform (CHCl3) supposed to be a carcinogen The level of contamination
was 15 ppm (by mass):
(i) express this in percent by mass
(ii) determine the molality of chloroform in the water sample
|
1
|
1613-1616
|
1 9
A sample of drinking water was found to be severely contaminated with
chloroform (CHCl3) supposed to be a carcinogen The level of contamination
was 15 ppm (by mass):
(i) express this in percent by mass
(ii) determine the molality of chloroform in the water sample 1
|
1
|
1614-1617
|
9
A sample of drinking water was found to be severely contaminated with
chloroform (CHCl3) supposed to be a carcinogen The level of contamination
was 15 ppm (by mass):
(i) express this in percent by mass
(ii) determine the molality of chloroform in the water sample 1 10
What role does the molecular interaction play in a solution of alcohol and water
|
1
|
1615-1618
|
The level of contamination
was 15 ppm (by mass):
(i) express this in percent by mass
(ii) determine the molality of chloroform in the water sample 1 10
What role does the molecular interaction play in a solution of alcohol and water 1
|
1
|
1616-1619
|
1 10
What role does the molecular interaction play in a solution of alcohol and water 1 11
Why do gases always tend to be less soluble in liquids as the temperature
is raised
|
1
|
1617-1620
|
10
What role does the molecular interaction play in a solution of alcohol and water 1 11
Why do gases always tend to be less soluble in liquids as the temperature
is raised 1
|
1
|
1618-1621
|
1 11
Why do gases always tend to be less soluble in liquids as the temperature
is raised 1 12
State Henry’s law and mention some important applications
|
1
|
1619-1622
|
11
Why do gases always tend to be less soluble in liquids as the temperature
is raised 1 12
State Henry’s law and mention some important applications 1
|
1
|
1620-1623
|
1 12
State Henry’s law and mention some important applications 1 13
The partial pressure of ethane over a solution containing 6
|
1
|
1621-1624
|
12
State Henry’s law and mention some important applications 1 13
The partial pressure of ethane over a solution containing 6 56 × 10–3 g of ethane
is 1 bar
|
1
|
1622-1625
|
1 13
The partial pressure of ethane over a solution containing 6 56 × 10–3 g of ethane
is 1 bar If the solution contains 5
|
1
|
1623-1626
|
13
The partial pressure of ethane over a solution containing 6 56 × 10–3 g of ethane
is 1 bar If the solution contains 5 00 × 10–2 g of ethane, then what shall be the
partial pressure of the gas
|
1
|
1624-1627
|
56 × 10–3 g of ethane
is 1 bar If the solution contains 5 00 × 10–2 g of ethane, then what shall be the
partial pressure of the gas 1
|
1
|
1625-1628
|
If the solution contains 5 00 × 10–2 g of ethane, then what shall be the
partial pressure of the gas 1 14
What is meant by positive and negative deviations from Raoult's law and how is
the sign of DmixH related to positive and negative deviations from Raoult's law
|
1
|
1626-1629
|
00 × 10–2 g of ethane, then what shall be the
partial pressure of the gas 1 14
What is meant by positive and negative deviations from Raoult's law and how is
the sign of DmixH related to positive and negative deviations from Raoult's law 1
|
1
|
1627-1630
|
1 14
What is meant by positive and negative deviations from Raoult's law and how is
the sign of DmixH related to positive and negative deviations from Raoult's law 1 15
An aqueous solution of 2% non-volatile solute exerts a pressure of 1
|
1
|
1628-1631
|
14
What is meant by positive and negative deviations from Raoult's law and how is
the sign of DmixH related to positive and negative deviations from Raoult's law 1 15
An aqueous solution of 2% non-volatile solute exerts a pressure of 1 004 bar
at the normal boiling point of the solvent
|
1
|
1629-1632
|
1 15
An aqueous solution of 2% non-volatile solute exerts a pressure of 1 004 bar
at the normal boiling point of the solvent What is the molar mass of the solute
|
1
|
1630-1633
|
15
An aqueous solution of 2% non-volatile solute exerts a pressure of 1 004 bar
at the normal boiling point of the solvent What is the molar mass of the solute 1
|
1
|
1631-1634
|
004 bar
at the normal boiling point of the solvent What is the molar mass of the solute 1 16
Heptane and octane form an ideal solution
|
1
|
1632-1635
|
What is the molar mass of the solute 1 16
Heptane and octane form an ideal solution At 373 K, the vapour pressures of
the two liquid components are 105
|
1
|
1633-1636
|
1 16
Heptane and octane form an ideal solution At 373 K, the vapour pressures of
the two liquid components are 105 2 kPa and 46
|
1
|
1634-1637
|
16
Heptane and octane form an ideal solution At 373 K, the vapour pressures of
the two liquid components are 105 2 kPa and 46 8 kPa respectively
|
1
|
1635-1638
|
At 373 K, the vapour pressures of
the two liquid components are 105 2 kPa and 46 8 kPa respectively What will
be the vapour pressure of a mixture of 26
|
1
|
1636-1639
|
2 kPa and 46 8 kPa respectively What will
be the vapour pressure of a mixture of 26 0 g of heptane and 35 g of octane
|
1
|
1637-1640
|
8 kPa respectively What will
be the vapour pressure of a mixture of 26 0 g of heptane and 35 g of octane 1
|
1
|
1638-1641
|
What will
be the vapour pressure of a mixture of 26 0 g of heptane and 35 g of octane 1 17
The vapour pressure of water is 12
|
1
|
1639-1642
|
0 g of heptane and 35 g of octane 1 17
The vapour pressure of water is 12 3 kPa at 300 K
|
1
|
1640-1643
|
1 17
The vapour pressure of water is 12 3 kPa at 300 K Calculate vapour pressure
of 1 molal solution of a non-volatile solute in it
|
1
|
1641-1644
|
17
The vapour pressure of water is 12 3 kPa at 300 K Calculate vapour pressure
of 1 molal solution of a non-volatile solute in it 1
|
1
|
1642-1645
|
3 kPa at 300 K Calculate vapour pressure
of 1 molal solution of a non-volatile solute in it 1 18
Calculate the mass of a non-volatile solute (molar mass 40 g mol–1) which
should be dissolved in 114 g octane to reduce its vapour pressure to 80%
|
1
|
1643-1646
|
Calculate vapour pressure
of 1 molal solution of a non-volatile solute in it 1 18
Calculate the mass of a non-volatile solute (molar mass 40 g mol–1) which
should be dissolved in 114 g octane to reduce its vapour pressure to 80% 1
|
1
|
1644-1647
|
1 18
Calculate the mass of a non-volatile solute (molar mass 40 g mol–1) which
should be dissolved in 114 g octane to reduce its vapour pressure to 80% 1 19
A solution containing 30 g of non-volatile solute exactly in 90 g of water has a
vapour pressure of 2
|
1
|
1645-1648
|
18
Calculate the mass of a non-volatile solute (molar mass 40 g mol–1) which
should be dissolved in 114 g octane to reduce its vapour pressure to 80% 1 19
A solution containing 30 g of non-volatile solute exactly in 90 g of water has a
vapour pressure of 2 8 kPa at 298 K
|
1
|
1646-1649
|
1 19
A solution containing 30 g of non-volatile solute exactly in 90 g of water has a
vapour pressure of 2 8 kPa at 298 K Further, 18 g of water is then added to
the solution and the new vapour pressure becomes 2
|
1
|
1647-1650
|
19
A solution containing 30 g of non-volatile solute exactly in 90 g of water has a
vapour pressure of 2 8 kPa at 298 K Further, 18 g of water is then added to
the solution and the new vapour pressure becomes 2 9 kPa at 298 K
|
1
|
1648-1651
|
8 kPa at 298 K Further, 18 g of water is then added to
the solution and the new vapour pressure becomes 2 9 kPa at 298 K Calculate:
(i) molar mass of the solute
(ii) vapour pressure of water at 298 K
|
1
|
1649-1652
|
Further, 18 g of water is then added to
the solution and the new vapour pressure becomes 2 9 kPa at 298 K Calculate:
(i) molar mass of the solute
(ii) vapour pressure of water at 298 K 1
|
1
|
1650-1653
|
9 kPa at 298 K Calculate:
(i) molar mass of the solute
(ii) vapour pressure of water at 298 K 1 20
A 5% solution (by mass) of cane sugar in water has freezing point of 271K
|
1
|
1651-1654
|
Calculate:
(i) molar mass of the solute
(ii) vapour pressure of water at 298 K 1 20
A 5% solution (by mass) of cane sugar in water has freezing point of 271K Calculate the freezing point of 5% glucose in water if freezing point of pure
water is 273
|
1
|
1652-1655
|
1 20
A 5% solution (by mass) of cane sugar in water has freezing point of 271K Calculate the freezing point of 5% glucose in water if freezing point of pure
water is 273 15 K
|
1
|
1653-1656
|
20
A 5% solution (by mass) of cane sugar in water has freezing point of 271K Calculate the freezing point of 5% glucose in water if freezing point of pure
water is 273 15 K 1
|
1
|
1654-1657
|
Calculate the freezing point of 5% glucose in water if freezing point of pure
water is 273 15 K 1 21
Two elements A and B form compounds having formula AB2 and AB4
|
1
|
1655-1658
|
15 K 1 21
Two elements A and B form compounds having formula AB2 and AB4 When
dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by
2
|
1
|
1656-1659
|
1 21
Two elements A and B form compounds having formula AB2 and AB4 When
dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by
2 3 K whereas 1
|
1
|
1657-1660
|
21
Two elements A and B form compounds having formula AB2 and AB4 When
dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by
2 3 K whereas 1 0 g of AB4 lowers it by 1
|
1
|
1658-1661
|
When
dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by
2 3 K whereas 1 0 g of AB4 lowers it by 1 3 K
|
1
|
1659-1662
|
3 K whereas 1 0 g of AB4 lowers it by 1 3 K The molar depression constant
for benzene is 5
|
1
|
1660-1663
|
0 g of AB4 lowers it by 1 3 K The molar depression constant
for benzene is 5 1 K kg mol–1
|
1
|
1661-1664
|
3 K The molar depression constant
for benzene is 5 1 K kg mol–1 Calculate atomic masses of A and B
|
1
|
1662-1665
|
The molar depression constant
for benzene is 5 1 K kg mol–1 Calculate atomic masses of A and B Rationalised 2023-24
29
Solutions
1
|
1
|
1663-1666
|
1 K kg mol–1 Calculate atomic masses of A and B Rationalised 2023-24
29
Solutions
1 22
At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure
of 4
|
1
|
1664-1667
|
Calculate atomic masses of A and B Rationalised 2023-24
29
Solutions
1 22
At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure
of 4 98 bar
|
1
|
1665-1668
|
Rationalised 2023-24
29
Solutions
1 22
At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure
of 4 98 bar If the osmotic pressure of the solution is 1
|
1
|
1666-1669
|
22
At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure
of 4 98 bar If the osmotic pressure of the solution is 1 52 bars at the same
temperature, what would be its concentration
|
1
|
1667-1670
|
98 bar If the osmotic pressure of the solution is 1 52 bars at the same
temperature, what would be its concentration 1
|
1
|
1668-1671
|
If the osmotic pressure of the solution is 1 52 bars at the same
temperature, what would be its concentration 1 23
Suggest the most important type of intermolecular attractive interaction in
the following pairs
|
1
|
1669-1672
|
52 bars at the same
temperature, what would be its concentration 1 23
Suggest the most important type of intermolecular attractive interaction in
the following pairs (i) n-hexane and n-octane
(ii) I2 and CCl4
(iii) NaClO4 and water
(iv) methanol and acetone
(v) acetonitrile (CH3CN) and acetone (C3H6O)
|
1
|
1670-1673
|
1 23
Suggest the most important type of intermolecular attractive interaction in
the following pairs (i) n-hexane and n-octane
(ii) I2 and CCl4
(iii) NaClO4 and water
(iv) methanol and acetone
(v) acetonitrile (CH3CN) and acetone (C3H6O) 1
|
1
|
1671-1674
|
23
Suggest the most important type of intermolecular attractive interaction in
the following pairs (i) n-hexane and n-octane
(ii) I2 and CCl4
(iii) NaClO4 and water
(iv) methanol and acetone
(v) acetonitrile (CH3CN) and acetone (C3H6O) 1 24
Based on solute-solvent interactions, arrange the following in order of increasing
solubility in n-octane and explain
|
1
|
1672-1675
|
(i) n-hexane and n-octane
(ii) I2 and CCl4
(iii) NaClO4 and water
(iv) methanol and acetone
(v) acetonitrile (CH3CN) and acetone (C3H6O) 1 24
Based on solute-solvent interactions, arrange the following in order of increasing
solubility in n-octane and explain Cyclohexane, KCl, CH3OH, CH3CN
|
1
|
1673-1676
|
1 24
Based on solute-solvent interactions, arrange the following in order of increasing
solubility in n-octane and explain Cyclohexane, KCl, CH3OH, CH3CN 1
|
1
|
1674-1677
|
24
Based on solute-solvent interactions, arrange the following in order of increasing
solubility in n-octane and explain Cyclohexane, KCl, CH3OH, CH3CN 1 25
Amongst the following compounds, identify which are insoluble, partially
soluble and highly soluble in water
|
1
|
1675-1678
|
Cyclohexane, KCl, CH3OH, CH3CN 1 25
Amongst the following compounds, identify which are insoluble, partially
soluble and highly soluble in water (i) phenol
(ii) toluene
(iii) formic acid
(iv) ethylene glycol
(v) chloroform
(vi) pentanol
|
1
|
1676-1679
|
1 25
Amongst the following compounds, identify which are insoluble, partially
soluble and highly soluble in water (i) phenol
(ii) toluene
(iii) formic acid
(iv) ethylene glycol
(v) chloroform
(vi) pentanol 1
|
1
|
1677-1680
|
25
Amongst the following compounds, identify which are insoluble, partially
soluble and highly soluble in water (i) phenol
(ii) toluene
(iii) formic acid
(iv) ethylene glycol
(v) chloroform
(vi) pentanol 1 26
If the density of some lake water is 1
|
1
|
1678-1681
|
(i) phenol
(ii) toluene
(iii) formic acid
(iv) ethylene glycol
(v) chloroform
(vi) pentanol 1 26
If the density of some lake water is 1 25g mL–1 and contains 92 g of Na+ ions per
kg of water, calculate the molarity of Na+ ions in the lake
|
1
|
1679-1682
|
1 26
If the density of some lake water is 1 25g mL–1 and contains 92 g of Na+ ions per
kg of water, calculate the molarity of Na+ ions in the lake 1
|
1
|
1680-1683
|
26
If the density of some lake water is 1 25g mL–1 and contains 92 g of Na+ ions per
kg of water, calculate the molarity of Na+ ions in the lake 1 27
If the solubility product of CuS is 6 × 10–16, calculate the maximum molarity of
CuS in aqueous solution
|
1
|
1681-1684
|
25g mL–1 and contains 92 g of Na+ ions per
kg of water, calculate the molarity of Na+ ions in the lake 1 27
If the solubility product of CuS is 6 × 10–16, calculate the maximum molarity of
CuS in aqueous solution 1
|
1
|
1682-1685
|
1 27
If the solubility product of CuS is 6 × 10–16, calculate the maximum molarity of
CuS in aqueous solution 1 28
Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when
6
|
1
|
1683-1686
|
27
If the solubility product of CuS is 6 × 10–16, calculate the maximum molarity of
CuS in aqueous solution 1 28
Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when
6 5 g of C9H8O4 is dissolved in 450 g of CH3CN
|
1
|
1684-1687
|
1 28
Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when
6 5 g of C9H8O4 is dissolved in 450 g of CH3CN 1
|
1
|
1685-1688
|
28
Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when
6 5 g of C9H8O4 is dissolved in 450 g of CH3CN 1 29
Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal
symptoms in narcotic users
|
1
|
1686-1689
|
5 g of C9H8O4 is dissolved in 450 g of CH3CN 1 29
Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal
symptoms in narcotic users Dose of nalorphene generally given is 1
|
1
|
1687-1690
|
1 29
Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal
symptoms in narcotic users Dose of nalorphene generally given is 1 5 mg
|
1
|
1688-1691
|
29
Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal
symptoms in narcotic users Dose of nalorphene generally given is 1 5 mg Calculate the mass of 1
|
1
|
1689-1692
|
Dose of nalorphene generally given is 1 5 mg Calculate the mass of 1 5 ´ 10–3 m aqueous solution required for the above dose
|
1
|
1690-1693
|
5 mg Calculate the mass of 1 5 ´ 10–3 m aqueous solution required for the above dose 1
|
1
|
1691-1694
|
Calculate the mass of 1 5 ´ 10–3 m aqueous solution required for the above dose 1 30
Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250
mL of 0
|
1
|
1692-1695
|
5 ´ 10–3 m aqueous solution required for the above dose 1 30
Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250
mL of 0 15 M solution in methanol
|
1
|
1693-1696
|
1 30
Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250
mL of 0 15 M solution in methanol 1
|
1
|
1694-1697
|
30
Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250
mL of 0 15 M solution in methanol 1 31
The depression in freezing point of water observed for the same amount of
acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order
given above
|
1
|
1695-1698
|
15 M solution in methanol 1 31
The depression in freezing point of water observed for the same amount of
acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order
given above Explain briefly
|
1
|
1696-1699
|
1 31
The depression in freezing point of water observed for the same amount of
acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order
given above Explain briefly 1
|
1
|
1697-1700
|
31
The depression in freezing point of water observed for the same amount of
acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order
given above Explain briefly 1 32
Calculate the depression in the freezing point of water when 10 g of
CH3CH2CHClCOOH is added to 250 g of water
|
1
|
1698-1701
|
Explain briefly 1 32
Calculate the depression in the freezing point of water when 10 g of
CH3CH2CHClCOOH is added to 250 g of water Ka = 1
|
1
|
1699-1702
|
1 32
Calculate the depression in the freezing point of water when 10 g of
CH3CH2CHClCOOH is added to 250 g of water Ka = 1 4 × 10–3, Kf = 1
|
1
|
1700-1703
|
32
Calculate the depression in the freezing point of water when 10 g of
CH3CH2CHClCOOH is added to 250 g of water Ka = 1 4 × 10–3, Kf = 1 86
K kg mol–1
|
1
|
1701-1704
|
Ka = 1 4 × 10–3, Kf = 1 86
K kg mol–1 1
|
1
|
1702-1705
|
4 × 10–3, Kf = 1 86
K kg mol–1 1 33
19
|
1
|
1703-1706
|
86
K kg mol–1 1 33
19 5 g of CH2FCOOH is dissolved in 500 g of water
|
1
|
1704-1707
|
1 33
19 5 g of CH2FCOOH is dissolved in 500 g of water The depression in the freezing
point of water observed is 1
|
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