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www.thechemistryguru.com Methods of expressing Concentrations

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LIQUID SOLUTION

A solution is a homogenous mixture of two or more substances, the

composition of which may vary within limits. “A solution is a special

kind of mixture in which substances are intermixed so intimately that

they can not be observed as separate components”. The dispersed

phase or the substance which is to be dissolved is called solute, while

the dispersion medium in which the solute is dispersed to get a

homogenous mixture is called the solvent. A solution is termed as

binary, ternary and quartenary if it consists of two, three and four

components respectively.

Methods of expressing concentration of solution

Concentration of solution is the amount of solute dissolved in a

known amount of the solvent or solution. The concentration of

solution can be expressed in various ways as discussed below.

(1) Percentage : It refers to the amount of the solute per 100 parts of

the solution. It can also be called as parts per hundred (pph). It can

be expressed by any of following four methods :


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(i) Weight to weight percent (% w/w) Wt. of solute

100Wt. of solution

e.g., 10% 2 3Na CO solution w/w means 10g of 2 3Na CO is dissolved in

100g of the solution. (It means 10g 2 3Na CO is dissolved in 90g of

2H O)

(ii) Weight to volume percent (% w/v) Wt. of solute

100Volume of solution

e.g., 10% 2 3Na CO (w/v) means 10g 2 3Na CO is dissolved in

100 cc of solution.

(iii) Volume to volume percent (% v/v) Vol. of solute

100Vol. of solution

e.g., 10% ethanol (v/v) means 10 cc of ethanol dissolved in 100 cc of

solution.

(iv) Volume to weight percent (% v/w) Vol. of solute

100Wt. of solution

e.g., 10% ethanol (v/w) means 10 cc of ethanol dissolved in

g100 of solution.

(2) Parts per million (ppm) and parts per billion (ppb) : When a

solute is present in trace quantities, it is convenient to express the

concentration in parts per million and parts per billion. It is the

number of parts of solute per million )10( 6 or per billion 9(10 ) parts of

the solution. It is independent of the temperature.


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6mass of solute componentppm 10

Total mass of solution ;

9mass of solute componentppb 10

Total mass of solution

Example 1 : 15 g of methyl alcohol is dissolved in 35 g of water. The

weight percentage of methyl alcohol in solution is

(a) 30% (b) 50% (c) 70% (d) 75%

Solution (a) : Weight of solute

Weight percentage 100Weight of solution

Total weight of solution = (15 + 35) g = 50 g

Weight percentage of methyl alcohol

Weight of methyl alcohol 15100 100 30%

Weight of solution 50

Example 2 : Sea water contains 35.8 10 g of dissolved oxygen

per kilogram. The concentration of oxygen in parts per million is

(a) 5.8 ppm (b) 58.5ppm (c) 0.58ppm (d) 0.05ppm

Solution (a) :

Part per million 3

6 63

Mass of solute 5.8 10 g10 10

Mass of solution 10 g

5.8ppm

Examples based on percentage and parts per million (ppm)


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Example 3 : A 500gm toothpaste sample has g2.0 fluoride

concentration. The concentration of fluoride ions in terms of ppm

level is

(a) 250 ppm (b) 200 ppm (c) 400 ppm (d) 1000 ppm

Solution (c) :

ppm of F ions 6Mass of solute10

Mass of solution 60.2

10500

400 ppm

(3) Strength : The strength of solution is defined as the amount of

solute in grams present in one litre (or 3dm ) of the solution. It is

expressed in g/litre or 3(g / dm ) .

Mass of solute in gramsStrength

Volume of solution in litres

(4) Normality (N) : It is defined as the number of gram equivalents

(equivalent weight in grams) of a solute present per litre of the

solution. Unit of normality is gram equivalents litre–1. Normality

changes with temperature since it involves volume. When a solution

is diluted x times, its normality also decreases by x times. Solutions

in term of normality generally expressed as,

N Normal solution; 5N Penta normal, 10N Deca normal; N / 2

semi normal

N /10 Deci normal; N /5 Penti normal

N /100 or 0.01 N centinormal, N /1000 or 0.001= millinormal

Mathematically normality can be calculated by following formulas,


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(i) Number of g.eq. of solute

Normality (N)Volume of solution (l)

Weight of solute in g.

g. eq. weight of solute Volume of solution (l)

(ii) Wt. of solute per litre of solution

Ng eq. wt. of solute

,

(iii) Wt. of solute 1000

Ng.eq. wt. of solute Vol. of solution in ml

(iv) Percent of solute 10


,

(v) -1Strength in g l of solution


(vi) Wt% density 10

NEq. wt.

(vii) If volume 1V and normality 1N is so changed that new normality

and volume 2N and 2V then,

1 1 2 2N V N V (Normality equation)

(viii) When two solutions of the same solute are mixed then normality

of mixture )(N is

1 1 2 2

1 2

N V N VN

V V

(ix) Vol. of water to be added i.e., 2 1(V V ) to get a solution of

normality 2N from 1V ml of normality 1N


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1 22 1 1

2

N NV V V

N

(x) If Wg of an acid is completely neutralised by V ml of base of

normality N

Wt. of acid VN

g eq. wt. of acid 1000 ;Similarly,

Wt. of base Vol. of acid N of acid

g eq. wt. of base 1000

(xi) When aV ml of acid of normality aN is mixed with bV ml of base

of normality bN

(a) If a a b bV N V N (Solution neutral)

(b) If a a b bV N V N (Solution is acidic)

(c) If a ab bV N V N (Solution is basic)

(xii) Normality of the acidic mixture a a b b

a b

V N V N

(V V )

(xiii) Normality of the basic mixture a ab b

a b

V N V N

(V V )

(xiv) No. of meq * of solute

NVol. of solution in ml

(*1 equivalent = 1000 milliequivalent or meq.)


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Example 4 : Normality of a solution containing 9.8 g of 2 4H SO

in 3250 cm of the solution is

(a) 0.8 N (b) 1N (c) 0.08N (d) 1.8N

Solution (a) : Eq. wt. of 2 42 4

2 4

Mol. mass of H SOH SO

Basicity of H SO

9849

2

Number of g equivalent of 2 4H SO Weight in g 9.8

0.2Eq. mass 49

3250 cm of solution contain 0 0B B BA A AP P X ; P P X 0.2g equivalent

31000 cm of the solution contain 2 4H SO 0.2

1000g250

equivalent 0.8g equivalent

Hence normality of the solution 0.8N

Example 5 : Amount of NaOH present in 200 ml of 0.5N

solution is

(a) 40 g (b) 4 g (c) 0.4 g (d) 4.4 g

Solution (b) : Wt. of solute N V g eq. wt. 0.5 200 40

4g1000 1000

Examples based on Normality


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Example 6 : 50 ml of 2 410N H SO , 25ml of 12N HCl and

40 ml of 35N HNO were mixed together and the volume of the

mixture was made 1000ml by adding water. The normality of the

resulting solution will be

(a) 1N (b) 2N (c) 3N (d) 4N

Solution (a) : 1 1 2 2 3 3 4 4N V N V N V N V

450 10 25 12 40 5 N 1000 or 4N 1N

Example 7 : 100ml of 0.3N HCl is mixed with 200ml of

2 40.6N H SO . The final normality of the resulting solution will be

(a) 0.1 N (b) 0.2 N (c) 0.3N (d) 0.5N

Solution (d) : 100ml of 0.3N HCl contains HCl 0.03g eq.,

200ml of 2 40.6N H SO contains 2 4H SO 0.6

200 0.12g1000

eq.

Total g eq. 0.15 , Total volume 300ml Finally normality

0.151000 0.5

300

Alternatively 1 1 2 2 3 3N V N V N V i.e., 30.3 100 0.6 200 N 300 or

30.3 1.2 3N or 3N 1.5 /3 0.5


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Example 8 : An aqueous solution of 6.3g oxalic acid dihydrate is

made up to 250ml. The volume of 0.1N NaOH2

required to

completely neutralize 10ml. of this solution is

(a) 40 ml (b) 20 ml (c)10 ml (d) 4 ml

Solution (a) : Normality of oxalic acid solution 6.3 1000

0.463 250

1 1 2 2N V N V

10.1 V 0.4 10 or 1V 40 ml

Example 9 : 10.6 g of 2 3Na CO was exactly neutralised by

100 ml of 2 4H SO solution. Its normality is

(a) 1 N (b) 2 N (c) 1.5 N (d) 0.5 N

Solution (b) : Weight of base (w) 10.6g ; g eq. wt. of base = 53;

Vol. of acid (V) 100 ml ; Normality of acid (N) ?

w V N

g eq. wt. 1000

;

10.6 100 N

53 1000

;

1000 10.6N 2

100 53


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(4) Molarity (M) : Molarity of a solution is the number of moles of the

solute per litre of solution (or number of millimoles per ml. of

solution). Unit of molarity is mol/litre or mol/dm3 For example, a

molar 1M solution of sugar means a solution containing 1 mole of

sugar (i.e., 342 g or 236.02 10 molecules of it) per litre of the solution.

Solutions in term of molarity generally expressed as,

1M = Molar solution, M2 = Molarity is two,

M

2or 0.5 M = Semimolar solution,

M

10 or 0.1 M = Decimolar solution,

M

100 or 0.01 M = Centimolar solution

M

1000 or 0.001 M = Millimolar solution

Molarity is most common way of representing the concentration of

solution.

Molarity is depend on temperature as, 1

MolarityTemperature

When a solution is diluted (x times), its molarity also decreases (by

x times)


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Mathematically molarity can be calculated by following formulas,

(i) No. of moles of solute (n)

MVol. of solution in litres

,

(ii) Wt. of solute (in gm) per litre of solution

MMol. wt. of solute

(iii) Wt. of solute (in gm) 1000

MMol. wt. of solute Vol. of solution in ml.

(iv) No. of millimoles of solute

MVol. of solution in ml

(v) Percent of solute 10

MMol. wt. of solute

(vi) -1Strength in gl of solution

MMol. wt. of solute

(vii) 10 Sp. gr. of the solution Wt. % of the solute

MMol. wt. of the solute

(viii) If molarity and volume of solution are changed from 1 1M ,V to

2 2M ,V . Then,

1 1 2 2M V M V (Molarity equation)

(ix) In balanced chemical equation, if 1n moles of reactant one react

with 2n moles of reactant two. Then,

1 1 2 2

1 2

M V M V

n n


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(x) If two solutions of the same solute are mixed then molarity (M)

of resulting solution.

1 1 2 2

1 2

M V M VM

(V V )

(xi) Volume of water added to get a solution of molarity 2M from

1V ml of molarity 1M is

1 22 1 1

2

M MV V V

M

Relation between molarity and normality

Normality of solution = molarity Molecular mass

Equivalent mass

Normality equivalent mass = molarity molecular mass

For an acid, Molecular mass

Equivalent mass= basicity

So, Normality of acid = molarity basicity.

For a base, Molecular mass

Equivalent mass= Acidity

So, Normality of base = Molarity Acidity.


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Example 10 : The molarity of pure water (d 1 g / l) is

(a) 555 M (b) 5.55 M (c) 55.5 M (d) None

Solution (c) : Consider 1000ml of water

Mass of 100 ml of water 1000 1 1000 g

Number of moles of water 1000

55.518

Molarity No. of moles of water 55.5

55.5MVolume in litre 1

Example 11 : The number of iodine atoms present in 31cm of its

0.1M solution is

(a) 236.02 10 (b) 226.02 10 (c) 196.02 10 (d) 201.204 10

Solution (d) :

1 cc of 20.1M I solution 40.110 mol

1000 4 2310 6.02 10

molecules 196.02 10 molecules

19 202 6.02 10 atoms 1.204 10

Examples based on Molarity


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Example 12 : Equal volumes of 30.1M AgNO and 0.2M NaCl

are mixed. The concentration of 3NO ions in the mixture will be

(a) 0.1M (b) 0.05 M (c) 0.2 M (d) 0.15 M

Solution (b) : 3 30.2M0.1M

AgNO NaCl AgCl NaNO

30.1M AgNO reacts with 0.1M NaCl to produce

0.1M AgCl and 30.1M NaNO

3

0.1MNO 0.05M

2

[ when equal volumes are mixed dilution occurs]

Example 13 : The molarity of 2 4H SO solution that has a density

of 1.84g / cc at o35 C and contains 98% by weight is

(a) 4.18 M (b) 8.14 M (c) 18.4 M (d) 18 M

Solution (c) : Molarity

Wt. of solute 1000

Mol. wt. Vol. of solution (in ml.)

98 100018.4M

98 54.34

mass 100

Vol. of solution 54.34 mldensity 1.84


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Example 14 : Amount of oxalic acid 2 2(COOH) .2H O) in grams

that is required to obtain 250 ml of a semi-molar solution is

(a) 17.25 g (b) 17.00 g (c) 15.75 g (d) 15.00 g

Solution (c) : Molecular mass of oxalic acid = 126

1000ml of 1 M oxalic acid require oxalic acid =126 g

250 ml of 1 M oxalic acid will require oxalic acid

126250 31.5g

1000

Hence 250ml of 2

M oxalic acid will require oxalic acid

131.5 15.75 g

2

Mass of oxalic acid required 15.75g .

Example 15 : Volume of 10M HCl should be diluted with water

to prepare 2.00L of 5M HCl is

(a)2L (b) 1.5L (c) 1.00L (d)0.5L

Solution (c) : In dilution, the following equation is applicable :

1 1M V 2 2M V

10M HCl 5M HCl

110 V 5 2.00

1

5 2.00V L 1.00L

10


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Example 16 : The volume of 95% 2 4H SO (density 31.85g cm )

needed to prepare 3100cm of 15% solution of 2 4H SO (density =

31.10g cm ) will be

(a) 5 cc (b) 7.5 cc (c) 9.4 cc (d) 12.4cc

Solution (c) : Molarity of 95% 2 4

95 1H SO 1000

98 100 /1.85

17.93M

Molarity of 15% 2 4

15 1H SO 1000

98 100 /1.10 1.68 M

1 1M V = 2 2M V

(95% 2 4H SO ) (15% 2 4H SO )

117.93 V 1.68 100 or 2 5C H OH

Example 17 : The molarity of a 2 30.2N Na CO solution will be

(a) 0.05M (b) 0.2M (c) 0.1M (d) 0.4M

Solution (c) : 2molecular mass (M )N M

Equivalent mass (E)

So, 2

Equivalent mass (E)M N

Molecular mass (M )

Equivalent mass of salt Molecular mass

Total positive valency

Equivalent mass of 22 3

MNa CO

2


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2

2

M / 2M 0.2

M ;

0.2M

2 0.1M

Example 18 : If 20ml of 0.4N NaOH solution completely

neutralizes 40mL of dibasic acid. The molarity of acid solution is

(a) 0.1M (b) 0.2 M (b)0.3 M (d) 0.4 M

Solution (a) : We know 1 1 2 2N V N V

20.4 20 N 40 ; 2N 0.2N

Molarity 2CS ; Here acid solution is dibasic, so 0.2

M2

0.1M

(5) Molality (m) : It is the number of moles or gram molecules of the

solute per 1000 g of the solvent. Unit of molality is mol / kg . For

example, a 0.2 molal )2.0( m solution of glucose means a solution

obtained by dissolving 0.2 mole of glucose in gm1000 of water. Molality

(m) does not depend on temperature since it involves measurement of

weight of liquids. Molal solutions are less concentrated than molar

solution.

Mathematically molality can be calculated by following formulas,

(i) Number of moles of the solute

m 1000Weight of the solvent in grams

Strength per 1000 grams of solvent

Molecular mass of solute


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(ii) No. of gm moles of solute

mWt. of solvent in kg

(iii) Wt. of solute 1000

mMol. wt. of solute Wt. of solvent in g

(iv) No. of millimoles of solute

mWt. of solvent in g

(v) 10 solubility

mMol. wt. of solute

(vi) 1000 wt. % of solute (x)

m(100 x) mol. wt. of solute

(vii) 1000 Molarity

m(1000 sp. gravity) (Molarity Mol. wt. of solute)

Relation between molarity (M) and molality (m)

Molality (m) =Molarity

Molarity molecular massDensity

1000

Molarity (M) Molality density

Molality molecular mass1

1000


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Example 19 : 2 4H SO solution whose specific gravity is 11.98g ml

and 2 4H SO by volume is 95%. The molality of the solution will be

(a) 7.412 (b) 8.412 (c) 9.412 (d) 10.412

Solution (c) : 2 4H SO is 95% by volume

Wt. of 2 4H SO 95g Vol. of solution 100ml

moles of 2 4

95H SO

98 and weight of solution 100 1.98 198g

Weight of water 198 95 103 g

Molality 95 1000

9.41298 103

Hence molality of 2 4H SO solution is 9.412

Example 20 : The density of 2 4H SO solution is 11.84 gm ml . In

1 litre solution 2 4H SO is 93% by volume then, the molality of solution

is

(a) 9.42 (b) 10.42 (c) 11.42 (d) 12.42

Solution (b) : Given 2 4H SO is 93% by volume

Wt. of 2 4H SO 93 g

Examples based on Molality


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Volume of solution =100ml Density mass

volume mass d

volume

weight of solution 100 1.84 g 184 g

wt. of water 184 93 91 g

Molality Moles 93 1000

10.42wt. of water in kg 98 91

(6) Formality (F) : Formality of a solution may be defined as the

number of gram formula masses of the ionic solute dissolved per litre

of the solution. It is represented by F . Commonly, the term formality

is used to express the concentration of the ionic solids which do not

exist as molecules but exist as network of ions. A solution containing

one gram formula mass of solute per litre of the solution has

formality equal to one and is called formal solution. It may be

mentioned here that the formality of a solution changes with change

in temperature.

Formality

(F)=Number of gram formula masses of solute

Volume of solution in litres=

Mass of ionic solute (g)

(gm. formula mass of solute) (Volume of solution (l))

Thus, B BW (g) W (g) 1000F or

GFM V(l) GFM V(ml)


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(7) Mole fraction (X) : Mole fraction may be defined as the ratio of

number of moles of one component to the total number of moles of all

the components (solvent and solute) present in the solution. It is

denoted by the letter X . It may be noted that the mole fraction is

independent of the temperature. Mole fraction is dimensionless. Let us

suppose that a solution contains the components A and B and

suppose that AW g of A and BW g of B are present in it.

Number of moles of A is given by, AA

A

Wn

M and the number of moles

of B is given by, BB

B

Wn

M

where AM and BM are molecular masses of A and B respectively.

Total number of moles of A and BAB n n

Mole fraction of A , AA

BA

nX

n n

; Mole fraction of B , B

BBA

nX

n n

The sum of mole fractions of all the components in the solution is

always one.

A BBA

B BA A

n nX X 1

n n n n

.

Thus, if we know the mole fraction of one component of a binary

solution, the mole fraction of the other can be calculated.


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Relation between molality of solution (m) and mole fraction of

the solute (XA).

A

mX

55.5 m

Example 21 : A solution contains 16gm of methanol and gm10 of

water, mole fraction of methanol is

(a) 0.90 (b) 0.090 (c) 0.1 (d) 1.9

Solution (b) : Mass of methanol 16g , Mol. mass of 3CH OH 32

No. of moles of methanol 16

0.5 moles32

No. of moles of water =90

5 moles18

Mole fraction of methanol 0.5

0.0905 0.5

Example 22 : A solution has 25% of water, 25% ethanol and 50%

acetic acid by mass. The mole fraction of each component will be

(a) 0.50, 0.3, 0.19 (b) 0.19, 0.3, 0.50

(c) 0.3, 0.19, 0.5 (d) 0.50, 0.19, 0.3

Examples based on Mole fraction


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Solution (d) : Since 18g of water = 1mole

25g of water = 25

1.3818

mole

Similarly, g46 of ethanol = 1 mole

25g of ethanol 25

0.5546

moles

Again, 60g of acetic acid = 1 mole

50g of acetic acid 50

0.8360

mole

Mole fraction of water 1.38

0.501.38 0.55 0.83

Similarly, Mole fraction of ethanol 0.55

0.191.38 0.55 0.83

Mole fraction of acetic acid 0.83

0.31.38 0.55 0.83

(8) Mass fraction : Mass fraction of a component in a solution is the

mass of that component divided by the total mass of the solution. For

a solution containing Aw gm of A and Bw gm of B

A

BA

wMass fraction of A

w w

; B

BA

wMass fraction of B

w w


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Note : It may be noted that molality, mole fraction, mass fraction

etc. are preferred to molarity, normality, etc. because the former

involve the weights of the solute and solvent where as later involve

volumes of solutions. Temperature has no effect on weights but it has

significant effect on volumes.

(9) Demal unit (D) : The concentrations are also expressed in

“Demal unit”. One demal unit represents one mole of solute present

in one litre of solution at o0 C.

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