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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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