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1 | 605-608 | Concentration described by mass
percentage is commonly used in industrial chemical applications For example, commercial bleaching solution contains 3 62 mass
percentage of sodium hypochlorite in water (ii) Volume percentage (V/V): The volume percentage is defined as:
Volume % of a component =
Volume of the component
100
Total volume of solution
(1 |
1 | 606-609 | For example, commercial bleaching solution contains 3 62 mass
percentage of sodium hypochlorite in water (ii) Volume percentage (V/V): The volume percentage is defined as:
Volume % of a component =
Volume of the component
100
Total volume of solution
(1 2)
1 |
1 | 607-610 | 62 mass
percentage of sodium hypochlorite in water (ii) Volume percentage (V/V): The volume percentage is defined as:
Volume % of a component =
Volume of the component
100
Total volume of solution
(1 2)
1 2
1 |
1 | 608-611 | (ii) Volume percentage (V/V): The volume percentage is defined as:
Volume % of a component =
Volume of the component
100
Total volume of solution
(1 2)
1 2
1 2
1 |
1 | 609-612 | 2)
1 2
1 2
1 2
1 |
1 | 610-613 | 2
1 2
1 2
1 2
1 |
1 | 611-614 | 2
1 2
1 2
1 2
Expressing
Expressing
Expressing
Expressing
Expressing
Concentration
Concentration
Concentration
Concentration
Concentration
of Solutions
of Solutions
of Solutions
of Solutions
of Solutions
Rationalised 2023-24
3
Solutions
For example, 10% ethanol solution in water means that 10 mL
of ethanol is dissolved in water such that the total volume of
the solution is 100 mL |
1 | 612-615 | 2
1 2
1 2
Expressing
Expressing
Expressing
Expressing
Expressing
Concentration
Concentration
Concentration
Concentration
Concentration
of Solutions
of Solutions
of Solutions
of Solutions
of Solutions
Rationalised 2023-24
3
Solutions
For example, 10% ethanol solution in water means that 10 mL
of ethanol is dissolved in water such that the total volume of
the solution is 100 mL Solutions containing liquids are commonly
expressed in this unit |
1 | 613-616 | 2
1 2
Expressing
Expressing
Expressing
Expressing
Expressing
Concentration
Concentration
Concentration
Concentration
Concentration
of Solutions
of Solutions
of Solutions
of Solutions
of Solutions
Rationalised 2023-24
3
Solutions
For example, 10% ethanol solution in water means that 10 mL
of ethanol is dissolved in water such that the total volume of
the solution is 100 mL Solutions containing liquids are commonly
expressed in this unit For example, a 35% (v/v) solution of
ethylene glycol, an antifreeze, is used in cars for cooling the
engine |
1 | 614-617 | 2
Expressing
Expressing
Expressing
Expressing
Expressing
Concentration
Concentration
Concentration
Concentration
Concentration
of Solutions
of Solutions
of Solutions
of Solutions
of Solutions
Rationalised 2023-24
3
Solutions
For example, 10% ethanol solution in water means that 10 mL
of ethanol is dissolved in water such that the total volume of
the solution is 100 mL Solutions containing liquids are commonly
expressed in this unit For example, a 35% (v/v) solution of
ethylene glycol, an antifreeze, is used in cars for cooling the
engine At this concentration the antifreeze lowers the freezing
point of water to 255 |
1 | 615-618 | Solutions containing liquids are commonly
expressed in this unit For example, a 35% (v/v) solution of
ethylene glycol, an antifreeze, is used in cars for cooling the
engine At this concentration the antifreeze lowers the freezing
point of water to 255 4K (–17 |
1 | 616-619 | For example, a 35% (v/v) solution of
ethylene glycol, an antifreeze, is used in cars for cooling the
engine At this concentration the antifreeze lowers the freezing
point of water to 255 4K (–17 6°C) |
1 | 617-620 | At this concentration the antifreeze lowers the freezing
point of water to 255 4K (–17 6°C) (iii) Mass by volume percentage (w/V): Another unit which is
commonly used in medicine and pharmacy is mass by
volume percentage |
1 | 618-621 | 4K (–17 6°C) (iii) Mass by volume percentage (w/V): Another unit which is
commonly used in medicine and pharmacy is mass by
volume percentage It is the mass of solute dissolved in
100 mL of the solution |
1 | 619-622 | 6°C) (iii) Mass by volume percentage (w/V): Another unit which is
commonly used in medicine and pharmacy is mass by
volume percentage It is the mass of solute dissolved in
100 mL of the solution (iv) Parts per million: When a solute is present in trace quantities, it
is convenient to express concentration in parts per million (ppm)
and is defined as:
Parts per million =
6
Number of parts of the component
×10
Total number of parts of all components of the solution
(1 |
1 | 620-623 | (iii) Mass by volume percentage (w/V): Another unit which is
commonly used in medicine and pharmacy is mass by
volume percentage It is the mass of solute dissolved in
100 mL of the solution (iv) Parts per million: When a solute is present in trace quantities, it
is convenient to express concentration in parts per million (ppm)
and is defined as:
Parts per million =
6
Number of parts of the component
×10
Total number of parts of all components of the solution
(1 3)
As in the case of percentage, concentration in parts per million can
also be expressed as mass to mass, volume to volume and mass to
volume |
1 | 621-624 | It is the mass of solute dissolved in
100 mL of the solution (iv) Parts per million: When a solute is present in trace quantities, it
is convenient to express concentration in parts per million (ppm)
and is defined as:
Parts per million =
6
Number of parts of the component
×10
Total number of parts of all components of the solution
(1 3)
As in the case of percentage, concentration in parts per million can
also be expressed as mass to mass, volume to volume and mass to
volume A litre of sea water (which weighs 1030 g) contains about
6 × 10
–3 g of dissolved oxygen (O2) |
1 | 622-625 | (iv) Parts per million: When a solute is present in trace quantities, it
is convenient to express concentration in parts per million (ppm)
and is defined as:
Parts per million =
6
Number of parts of the component
×10
Total number of parts of all components of the solution
(1 3)
As in the case of percentage, concentration in parts per million can
also be expressed as mass to mass, volume to volume and mass to
volume A litre of sea water (which weighs 1030 g) contains about
6 × 10
–3 g of dissolved oxygen (O2) Such a small concentration is
also expressed as 5 |
1 | 623-626 | 3)
As in the case of percentage, concentration in parts per million can
also be expressed as mass to mass, volume to volume and mass to
volume A litre of sea water (which weighs 1030 g) contains about
6 × 10
–3 g of dissolved oxygen (O2) Such a small concentration is
also expressed as 5 8 g per 10
6 g (5 |
1 | 624-627 | A litre of sea water (which weighs 1030 g) contains about
6 × 10
–3 g of dissolved oxygen (O2) Such a small concentration is
also expressed as 5 8 g per 10
6 g (5 8 ppm) of sea water |
1 | 625-628 | Such a small concentration is
also expressed as 5 8 g per 10
6 g (5 8 ppm) of sea water The
concentration of pollutants in water or atmosphere is often expressed
in terms of mg mL
–1 or ppm |
1 | 626-629 | 8 g per 10
6 g (5 8 ppm) of sea water The
concentration of pollutants in water or atmosphere is often expressed
in terms of mg mL
–1 or ppm (v) Mole fraction: Commonly used symbol for mole fraction is x and
subscript used on the right hand side of x denotes the component |
1 | 627-630 | 8 ppm) of sea water The
concentration of pollutants in water or atmosphere is often expressed
in terms of mg mL
–1 or ppm (v) Mole fraction: Commonly used symbol for mole fraction is x and
subscript used on the right hand side of x denotes the component It is defined as:
Mole fraction of a component =
Number of moles of the component
Total number of moles of all the components
(1 |
1 | 628-631 | The
concentration of pollutants in water or atmosphere is often expressed
in terms of mg mL
–1 or ppm (v) Mole fraction: Commonly used symbol for mole fraction is x and
subscript used on the right hand side of x denotes the component It is defined as:
Mole fraction of a component =
Number of moles of the component
Total number of moles of all the components
(1 4)
For example, in a binary mixture, if the number of moles of A and B are
nA and nB respectively, the mole fraction of A will be
xA =
A
A
B
n
n
n
(1 |
1 | 629-632 | (v) Mole fraction: Commonly used symbol for mole fraction is x and
subscript used on the right hand side of x denotes the component It is defined as:
Mole fraction of a component =
Number of moles of the component
Total number of moles of all the components
(1 4)
For example, in a binary mixture, if the number of moles of A and B are
nA and nB respectively, the mole fraction of A will be
xA =
A
A
B
n
n
n
(1 5)
For a solution containing i number of components, we have:
xi =
i
1
2
i |
1 | 630-633 | It is defined as:
Mole fraction of a component =
Number of moles of the component
Total number of moles of all the components
(1 4)
For example, in a binary mixture, if the number of moles of A and B are
nA and nB respectively, the mole fraction of A will be
xA =
A
A
B
n
n
n
(1 5)
For a solution containing i number of components, we have:
xi =
i
1
2
i n
n
n
n =
i
i
n
n
(1 |
1 | 631-634 | 4)
For example, in a binary mixture, if the number of moles of A and B are
nA and nB respectively, the mole fraction of A will be
xA =
A
A
B
n
n
n
(1 5)
For a solution containing i number of components, we have:
xi =
i
1
2
i n
n
n
n =
i
i
n
n
(1 6)
It can be shown that in a given solution sum of all the mole
fractions is unity, i |
1 | 632-635 | 5)
For a solution containing i number of components, we have:
xi =
i
1
2
i n
n
n
n =
i
i
n
n
(1 6)
It can be shown that in a given solution sum of all the mole
fractions is unity, i e |
1 | 633-636 | n
n
n
n =
i
i
n
n
(1 6)
It can be shown that in a given solution sum of all the mole
fractions is unity, i e x1 + x2 + |
1 | 634-637 | 6)
It can be shown that in a given solution sum of all the mole
fractions is unity, i e x1 + x2 + + xi = 1
(1 |
1 | 635-638 | e x1 + x2 + + xi = 1
(1 7)
Mole fraction unit is very useful in relating some physical properties
of solutions, say vapour pressure with the concentration of the
solution and quite useful in describing the calculations involving
gas mixtures |
1 | 636-639 | x1 + x2 + + xi = 1
(1 7)
Mole fraction unit is very useful in relating some physical properties
of solutions, say vapour pressure with the concentration of the
solution and quite useful in describing the calculations involving
gas mixtures Rationalised 2023-24
4
Chemistry
Calculate the mole fraction of ethylene glycol (C2H6O2) in a solution
containing 20% of C2H6O2 by mass |
1 | 637-640 | + xi = 1
(1 7)
Mole fraction unit is very useful in relating some physical properties
of solutions, say vapour pressure with the concentration of the
solution and quite useful in describing the calculations involving
gas mixtures Rationalised 2023-24
4
Chemistry
Calculate the mole fraction of ethylene glycol (C2H6O2) in a solution
containing 20% of C2H6O2 by mass Assume that we have 100 g of solution (one can start with any amount of
solution because the results obtained will be the same) |
1 | 638-641 | 7)
Mole fraction unit is very useful in relating some physical properties
of solutions, say vapour pressure with the concentration of the
solution and quite useful in describing the calculations involving
gas mixtures Rationalised 2023-24
4
Chemistry
Calculate the mole fraction of ethylene glycol (C2H6O2) in a solution
containing 20% of C2H6O2 by mass Assume that we have 100 g of solution (one can start with any amount of
solution because the results obtained will be the same) Solution will
contain 20 g of ethylene glycol and 80 g of water |
1 | 639-642 | Rationalised 2023-24
4
Chemistry
Calculate the mole fraction of ethylene glycol (C2H6O2) in a solution
containing 20% of C2H6O2 by mass Assume that we have 100 g of solution (one can start with any amount of
solution because the results obtained will be the same) Solution will
contain 20 g of ethylene glycol and 80 g of water Molar mass of C2H6O2 = 12 × 2 + 1 × 6 + 16 × 2 = 62 g mol–1 |
1 | 640-643 | Assume that we have 100 g of solution (one can start with any amount of
solution because the results obtained will be the same) Solution will
contain 20 g of ethylene glycol and 80 g of water Molar mass of C2H6O2 = 12 × 2 + 1 × 6 + 16 × 2 = 62 g mol–1 Moles of C2H6O2 =
1
20 g
62 g mol
= 0 |
1 | 641-644 | Solution will
contain 20 g of ethylene glycol and 80 g of water Molar mass of C2H6O2 = 12 × 2 + 1 × 6 + 16 × 2 = 62 g mol–1 Moles of C2H6O2 =
1
20 g
62 g mol
= 0 322 mol
Moles of water =
-1
80 g
18 g mol
= 4 |
1 | 642-645 | Molar mass of C2H6O2 = 12 × 2 + 1 × 6 + 16 × 2 = 62 g mol–1 Moles of C2H6O2 =
1
20 g
62 g mol
= 0 322 mol
Moles of water =
-1
80 g
18 g mol
= 4 444 mol
2
6
2
glycol
2
6
2
2
moles of C H O
x
moles of C H O
moles of H O
0 |
1 | 643-646 | Moles of C2H6O2 =
1
20 g
62 g mol
= 0 322 mol
Moles of water =
-1
80 g
18 g mol
= 4 444 mol
2
6
2
glycol
2
6
2
2
moles of C H O
x
moles of C H O
moles of H O
0 322 mol
0 |
1 | 644-647 | 322 mol
Moles of water =
-1
80 g
18 g mol
= 4 444 mol
2
6
2
glycol
2
6
2
2
moles of C H O
x
moles of C H O
moles of H O
0 322 mol
0 322mol
4 |
1 | 645-648 | 444 mol
2
6
2
glycol
2
6
2
2
moles of C H O
x
moles of C H O
moles of H O
0 322 mol
0 322mol
4 444 mol = 0 |
1 | 646-649 | 322 mol
0 322mol
4 444 mol = 0 068
Similarly,
water
4 |
1 | 647-650 | 322mol
4 444 mol = 0 068
Similarly,
water
4 444 mol
0 |
1 | 648-651 | 444 mol = 0 068
Similarly,
water
4 444 mol
0 932
0 |
1 | 649-652 | 068
Similarly,
water
4 444 mol
0 932
0 322 mol
4 |
1 | 650-653 | 444 mol
0 932
0 322 mol
4 444 mol
x
Mole fraction of water can also be calculated as: 1 – 0 |
1 | 651-654 | 932
0 322 mol
4 444 mol
x
Mole fraction of water can also be calculated as: 1 – 0 068 = 0 |
1 | 652-655 | 322 mol
4 444 mol
x
Mole fraction of water can also be calculated as: 1 – 0 068 = 0 932
Example 1 |
1 | 653-656 | 444 mol
x
Mole fraction of water can also be calculated as: 1 – 0 068 = 0 932
Example 1 1
Example 1 |
1 | 654-657 | 068 = 0 932
Example 1 1
Example 1 1
Example 1 |
1 | 655-658 | 932
Example 1 1
Example 1 1
Example 1 1
Example 1 |
1 | 656-659 | 1
Example 1 1
Example 1 1
Example 1 1
Example 1 |
1 | 657-660 | 1
Example 1 1
Example 1 1
Example 1 1
(vi) Molarity: Molarity (M) is defined as number of moles of solute dissolved
in one litre (or one cubic decimetre) of solution,
Moles of solute
Molarity
Volume of solution in litre
(1 |
1 | 658-661 | 1
Example 1 1
Example 1 1
(vi) Molarity: Molarity (M) is defined as number of moles of solute dissolved
in one litre (or one cubic decimetre) of solution,
Moles of solute
Molarity
Volume of solution in litre
(1 8)
For example, 0 |
1 | 659-662 | 1
Example 1 1
(vi) Molarity: Molarity (M) is defined as number of moles of solute dissolved
in one litre (or one cubic decimetre) of solution,
Moles of solute
Molarity
Volume of solution in litre
(1 8)
For example, 0 25 mol L–1 (or 0 |
1 | 660-663 | 1
(vi) Molarity: Molarity (M) is defined as number of moles of solute dissolved
in one litre (or one cubic decimetre) of solution,
Moles of solute
Molarity
Volume of solution in litre
(1 8)
For example, 0 25 mol L–1 (or 0 25 M) solution of NaOH means that
0 |
1 | 661-664 | 8)
For example, 0 25 mol L–1 (or 0 25 M) solution of NaOH means that
0 25 mol of NaOH has been dissolved in one litre (or one cubic decimetre) |
1 | 662-665 | 25 mol L–1 (or 0 25 M) solution of NaOH means that
0 25 mol of NaOH has been dissolved in one litre (or one cubic decimetre) Example 1 |
1 | 663-666 | 25 M) solution of NaOH means that
0 25 mol of NaOH has been dissolved in one litre (or one cubic decimetre) Example 1 2
Example 1 |
1 | 664-667 | 25 mol of NaOH has been dissolved in one litre (or one cubic decimetre) Example 1 2
Example 1 2
Example 1 |
1 | 665-668 | Example 1 2
Example 1 2
Example 1 2
Example 1 |
1 | 666-669 | 2
Example 1 2
Example 1 2
Example 1 2
Example 1 |
1 | 667-670 | 2
Example 1 2
Example 1 2
Example 1 2
Calculate the molarity of a solution containing 5 g of NaOH in 450 mL
solution |
1 | 668-671 | 2
Example 1 2
Example 1 2
Calculate the molarity of a solution containing 5 g of NaOH in 450 mL
solution Moles of NaOH =
-1
5 g
40 g mol
= 0 |
1 | 669-672 | 2
Example 1 2
Calculate the molarity of a solution containing 5 g of NaOH in 450 mL
solution Moles of NaOH =
-1
5 g
40 g mol
= 0 125 mol
Volume of the solution in litres = 450 mL / 1000 mL L-1
Using equation (2 |
1 | 670-673 | 2
Calculate the molarity of a solution containing 5 g of NaOH in 450 mL
solution Moles of NaOH =
-1
5 g
40 g mol
= 0 125 mol
Volume of the solution in litres = 450 mL / 1000 mL L-1
Using equation (2 8),
Molarity =
–1
0 |
1 | 671-674 | Moles of NaOH =
-1
5 g
40 g mol
= 0 125 mol
Volume of the solution in litres = 450 mL / 1000 mL L-1
Using equation (2 8),
Molarity =
–1
0 125 mol × 1000 mL L
450 mL
= 0 |
1 | 672-675 | 125 mol
Volume of the solution in litres = 450 mL / 1000 mL L-1
Using equation (2 8),
Molarity =
–1
0 125 mol × 1000 mL L
450 mL
= 0 278 M
= 0 |
1 | 673-676 | 8),
Molarity =
–1
0 125 mol × 1000 mL L
450 mL
= 0 278 M
= 0 278 mol L–1
= 0 |
1 | 674-677 | 125 mol × 1000 mL L
450 mL
= 0 278 M
= 0 278 mol L–1
= 0 278 mol dm–3
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
5
Solutions
Example 1 |
1 | 675-678 | 278 M
= 0 278 mol L–1
= 0 278 mol dm–3
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
5
Solutions
Example 1 3
Example 1 |
1 | 676-679 | 278 mol L–1
= 0 278 mol dm–3
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
5
Solutions
Example 1 3
Example 1 3
Example 1 |
1 | 677-680 | 278 mol dm–3
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
5
Solutions
Example 1 3
Example 1 3
Example 1 3
Example 1 |
1 | 678-681 | 3
Example 1 3
Example 1 3
Example 1 3
Example 1 |
1 | 679-682 | 3
Example 1 3
Example 1 3
Example 1 3
Solution
Solution
Solution
Solution
Solution
(vii) Molality: Molality (m) is defined as the number of moles of the solute
per kilogram (kg) of the solvent and is expressed as:
Molality (m) =
Moles of solute
Mass of solvent in kg
(1 |
1 | 680-683 | 3
Example 1 3
Example 1 3
Solution
Solution
Solution
Solution
Solution
(vii) Molality: Molality (m) is defined as the number of moles of the solute
per kilogram (kg) of the solvent and is expressed as:
Molality (m) =
Moles of solute
Mass of solvent in kg
(1 9)
For example, 1 |
1 | 681-684 | 3
Example 1 3
Solution
Solution
Solution
Solution
Solution
(vii) Molality: Molality (m) is defined as the number of moles of the solute
per kilogram (kg) of the solvent and is expressed as:
Molality (m) =
Moles of solute
Mass of solvent in kg
(1 9)
For example, 1 00 mol kg
–1 (or 1 |
1 | 682-685 | 3
Solution
Solution
Solution
Solution
Solution
(vii) Molality: Molality (m) is defined as the number of moles of the solute
per kilogram (kg) of the solvent and is expressed as:
Molality (m) =
Moles of solute
Mass of solvent in kg
(1 9)
For example, 1 00 mol kg
–1 (or 1 00 m) solution of KCl means that
1 mol (74 |
1 | 683-686 | 9)
For example, 1 00 mol kg
–1 (or 1 00 m) solution of KCl means that
1 mol (74 5 g) of KCl is dissolved in 1 kg of water |
1 | 684-687 | 00 mol kg
–1 (or 1 00 m) solution of KCl means that
1 mol (74 5 g) of KCl is dissolved in 1 kg of water Each method of expressing concentration of the solutions has its
own merits and demerits |
1 | 685-688 | 00 m) solution of KCl means that
1 mol (74 5 g) of KCl is dissolved in 1 kg of water Each method of expressing concentration of the solutions has its
own merits and demerits Mass %, ppm, mole fraction and molality
are independent of temperature, whereas molarity is a function of
temperature |
1 | 686-689 | 5 g) of KCl is dissolved in 1 kg of water Each method of expressing concentration of the solutions has its
own merits and demerits Mass %, ppm, mole fraction and molality
are independent of temperature, whereas molarity is a function of
temperature This is because volume depends on temperature
and the mass does not |
1 | 687-690 | Each method of expressing concentration of the solutions has its
own merits and demerits Mass %, ppm, mole fraction and molality
are independent of temperature, whereas molarity is a function of
temperature This is because volume depends on temperature
and the mass does not Solubility of a substance is its maximum amount that can be dissolved
in a specified amount of solvent at a specified temperature |
1 | 688-691 | Mass %, ppm, mole fraction and molality
are independent of temperature, whereas molarity is a function of
temperature This is because volume depends on temperature
and the mass does not Solubility of a substance is its maximum amount that can be dissolved
in a specified amount of solvent at a specified temperature It depends
upon the nature of solute and solvent as well as temperature and
pressure |
1 | 689-692 | This is because volume depends on temperature
and the mass does not Solubility of a substance is its maximum amount that can be dissolved
in a specified amount of solvent at a specified temperature It depends
upon the nature of solute and solvent as well as temperature and
pressure Let us consider the effect of these factors in solution of a solid
or a gas in a liquid |
1 | 690-693 | Solubility of a substance is its maximum amount that can be dissolved
in a specified amount of solvent at a specified temperature It depends
upon the nature of solute and solvent as well as temperature and
pressure Let us consider the effect of these factors in solution of a solid
or a gas in a liquid 1 |
1 | 691-694 | It depends
upon the nature of solute and solvent as well as temperature and
pressure Let us consider the effect of these factors in solution of a solid
or a gas in a liquid 1 3 Solubility
1 |
1 | 692-695 | Let us consider the effect of these factors in solution of a solid
or a gas in a liquid 1 3 Solubility
1 3 Solubility
1 |
1 | 693-696 | 1 3 Solubility
1 3 Solubility
1 3 Solubility
1 |
1 | 694-697 | 3 Solubility
1 3 Solubility
1 3 Solubility
1 3 Solubility
1 |
1 | 695-698 | 3 Solubility
1 3 Solubility
1 3 Solubility
1 3 Solubility
Calculate molality of 2 |
1 | 696-699 | 3 Solubility
1 3 Solubility
1 3 Solubility
Calculate molality of 2 5 g of ethanoic acid (CH3COOH) in 75 g of benzene |
1 | 697-700 | 3 Solubility
1 3 Solubility
Calculate molality of 2 5 g of ethanoic acid (CH3COOH) in 75 g of benzene Molar mass of C2H4O2: 12 × 2 + 1 × 4 + 16 × 2 = 60 g mol–1
Moles of C2H4O2 =
1
2 |
1 | 698-701 | 3 Solubility
Calculate molality of 2 5 g of ethanoic acid (CH3COOH) in 75 g of benzene Molar mass of C2H4O2: 12 × 2 + 1 × 4 + 16 × 2 = 60 g mol–1
Moles of C2H4O2 =
1
2 5 g
60 g mol− = 0 |
1 | 699-702 | 5 g of ethanoic acid (CH3COOH) in 75 g of benzene Molar mass of C2H4O2: 12 × 2 + 1 × 4 + 16 × 2 = 60 g mol–1
Moles of C2H4O2 =
1
2 5 g
60 g mol− = 0 0417 mol
Mass of benzene in kg = 75 g/1000 g kg–1 = 75 × 10–3 kg
Molality of C2H4O2 =
2
4
2
Moles of C H O
kg of benzene
=
1
0 |
1 | 700-703 | Molar mass of C2H4O2: 12 × 2 + 1 × 4 + 16 × 2 = 60 g mol–1
Moles of C2H4O2 =
1
2 5 g
60 g mol− = 0 0417 mol
Mass of benzene in kg = 75 g/1000 g kg–1 = 75 × 10–3 kg
Molality of C2H4O2 =
2
4
2
Moles of C H O
kg of benzene
=
1
0 0417 mol
1000 g kg
75 g
−
×
= 0 |
1 | 701-704 | 5 g
60 g mol− = 0 0417 mol
Mass of benzene in kg = 75 g/1000 g kg–1 = 75 × 10–3 kg
Molality of C2H4O2 =
2
4
2
Moles of C H O
kg of benzene
=
1
0 0417 mol
1000 g kg
75 g
−
×
= 0 556 mol kg–1
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1 |
1 | 702-705 | 0417 mol
Mass of benzene in kg = 75 g/1000 g kg–1 = 75 × 10–3 kg
Molality of C2H4O2 =
2
4
2
Moles of C H O
kg of benzene
=
1
0 0417 mol
1000 g kg
75 g
−
×
= 0 556 mol kg–1
Intext Questions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
1 1 Calculate the mass percentage of benzene (C6H6) and carbon
tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of
carbon tetrachloride |
1 | 703-706 | 0417 mol
1000 g kg
75 g
−
×
= 0 556 mol kg–1
Intext Questions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
1 1 Calculate the mass percentage of benzene (C6H6) and carbon
tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of
carbon tetrachloride 1 |
1 | 704-707 | 556 mol kg–1
Intext Questions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
1 1 Calculate the mass percentage of benzene (C6H6) and carbon
tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of
carbon tetrachloride 1 2 Calculate the mole fraction of benzene in solution containing 30%
by mass in carbon tetrachloride |
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