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Methods of Expressing the Concentration of a Solution The concentration of a solution can be expresses in a number of ways. The important methods are: Mass/Weight Percentage or Per cent by Mass/Weight : It is defined as the amount of solute in grams present in 100 grams of the solution.
The concentration of a solution can be expresses in a number of ways. The important methods are:
It is defined as the amount of solute in grams present in 100 grams of the solution.
The ratio mass of solute to the mass of solvent is termed as mass fraction.
Thus, Mass percentage of solute = Mass fraction × 100
10% solution of sugar by mass means that 10 grams of sugar is present in 100 grams of the solution, i.e., 10 grams of sugar has been dissolved in 90 grams of water. e
Question:
What is the weight percentage of urea solution in which 10 gm of urea is dissolved in 90 gm water.
Solution
Weight percentage of urea = (weight of urea/ weight of solution) 100
= 10/(90+10) 100 = 10% urea solution (w/W)
10% solution of HCl by volume means that 10 mL of liquid HCl is present in 100 mLof the solution.
A 10% mass by volume solution means that 10 gm solute is present in 100 mL of solution.
Before dilution After dilution
M_{1}V_{1 } = M_{2}V_{2}
Let there be three samples of solution (containing same solvent and solute) with their molarity M_{1}, M_{2}, M_{3} and volumes V_{1}, V_{2}, V_{3} respectively. These solutions are mixed; molarity of mixed solution may be given as:
M_{1}V_{1} + M_{2}V_{2} + M_{3}V_{3} = M_{R}(V_{1} + V_{2} + V_{3})
Where M_{R} = resultant molarity
V_{1} + V_{2} + V_{3} = resultant molarity
Molarity is dependent on volume; therefore, it depends on temperature.
1 M → Molar solution, i.e., molarity is 1
0.5 M or M/2 → Semimolar
0.1 M or M/10 → Decimolar
0.01 M or M/100 → Centimolar
0.001 M or M/1000 → Millimolar
3.65 gm of HCL gas is present in 100 mL of its aqueous solution.What is the molarity?
Molarity = = (3.65/36.5) 1000/100 = 1M
Molality of a solution is defined as the number of moles of solute dissolved in 1 Kg of the solvent.
Thus, if one gram molecule of a solute is present in 1 kg of the solvent, the concentration of the solution is said to be one molal.
Units of molarity: mol kg^{-1}???
Molality is the most convenient method to express the concentration because it involves the mass of liquids rather than their volumes. It is also independent of the variation in temperature.
Molality and solubity are related by the following relation.
Molality = Solubility×10/Molecular mass of the solute
[Solubility = Mass of solute in grams/Mass of solvent in grams × 100]
Let the density of the solution be d. Unit= g mL^{−1 }
Mass of solution = V × d
Mass of solute = number of moles × molecular mass of solute = n m_{A}
Mass of solvent, W = mass of solution – mass of solute = V × d – n × m_{A}
Thus,
Where m_{A} is molecular mass of solvent.
?
The density of a 3M sodium thiosulphate solution is 1.25 gm cm^{–3}. Calculate
i) the molalities of Na^{+} and S_{2}O_{3}^{2– } ions
ii) percentage of weight of solution
3 M Na_{2}S_{2}O_{3} (Sodium thiosulphate) solution means
3 moles Na_{2}S_{2}O_{3} is present in 1 L or, 1000 ml solution
Wt. of solute Na_{2}S_{2}O_{3} = 3×158
wt. of solution = v × d
= 1000 ml × 1.25 gm/ml
= 1000 × 1.25 gm
Wt. of solvent = (1000 × 1.25 – 3 ×158) gm H_{2}O
Molality = no. of moles of solute per 1000 gm solvent
= 3.865 mol kg^{–1} solvent
Now, Na_{2}S_{2}O_{3} 2Na^{+} +S_{2}O_{3}^{-}
a) Hence molality of Na^{+} = 2 ×3.865 mol kg^{-1}
= 7.73 mol kg^{–1}
Hence molality = 1× 3.865 mol kg^{–1}
= 3.865 mole/kg
b) % of wt. of solution
1 L i.e. 1000 ml solution containing 3 moles Na_{2}S_{2}O_{3}
1000 × 1.25 gm solution containing 3 ×158 gm Na_{2}S_{2}O_{3}
_{}
= 37.92%
The normality of a solution gives the number of gram equivalents of the solute present in one litre of the solution. ?
Number of gram equivalent of solute = Mass of solute in gram/ equivalent weight of solute
Equivalent weight of solute (E) = Molar mass of solute/ Valence factor
Valence factor for base = acidity of base
Valence factor for acid = basicity of acid
Valence factor for element = valency
1N = Normal = One gram equivalent of the solute per litre of solution = Normality is 1
N/2 = Seminormal = 0.5 g equivalent of the solute per litre of solution = Normality is 0.5
N/10 = Decinormal = 0.1 g equivalent of the solute per litre of solution = Normality is 0.1
N/100 = Centinormal = 0.01 g equivalent of the solute per litre of solution = Normality is 0.01
N/1000 = Millinormal = 0.001 g equivalent of the solute per litre of solution = Normality is 0.001
We know that
Molarity × Molecular mass = Strength of solution (g/L)
Similarly,
Normality × Equivalent mass = Normality of the solution (g/L)
Hence,
Molarity × Molecular mass = Normality × Equivalent mass
So, Normality = n × Molarity
__________________________________________
Let ‘d’ is the density of solution in g/mL
and
x is the percentage of the solute by mass.
Then,
The mole fraction of any component in a solution is the ratio of the number of moles of that component to the total number of moles of all components .
Total mole fraction of all the components of any solution is 1.
For a binary solution of A and B
And, X_{A}+X_{B} = 1
X_{A} = n/N+n and X_{B} = N/N+n
X_{A}/X_{B} = n/N = Moles of solute/Moles of solvent = w_{A}/m_{B}/w_{B}×m_{A}
X_{A}×1000/X_{B}×m_{B} = w_{A}×1000/w_{B}×m_{A} = m
or
When a solute is present in trace quantities, it is convenient to express concentration in parts per million (ppm)
In case of mass it may be expressed as : (Mass of solute/Mass of solution )× 106
In case of volume it may be expressed as: (Volume of solute/Volume of solution) × 106
So, concentration in parts per million can be expressed as mass to mass, volume to volume and mass to volume form.
Atmospheric pollution in cities is also expressed in ppm by volume. It refers to the volume of the pollutant in 10^{6} units of volume. 10 ppm of SO_{2} in air means 10 mL of SO_{2} is present in 10^{6} mL of air.
It is the number of formula mass in grams present per litre of solution. In case formula mass is equal to molecular mass, formality is equal to molarity. Like molarity and normality, the formality is also dependent on temperature. It is used for ionic compounds in which there is no existence of molecule. Mole of ionic compounds is called formole and molarity as formality.
Where,
w = weight of solute,
f = formula weight of solute
V= volume of solution
n_{f}_{ }= no. of gram formula weight
CH_{3}COOH exists as dimer in benzene. 1.2 gm of the acid was dissolved and the volume was made up to one litre using benzene. What is the formality?
Molar mass of CH_{3}COOH = 60
Formula weight of the associated molecule of the acid = 2 60 = 120
Weight of CH_{3}COOH = 1.2 gm
Volume of solution = 1 L
Formality = 1.2/120 1/1 = 0.01 F
Question 1: How much water would be present in 100 mL of 20% aqueous solution of sugar by volume?
a. 20 mL
b. 80 mL
c. 90 mL
d. 100 mL
Question 2: What would be concentration of the solution formed by adding two moles of solute in 1 kg water?
a. 2 M
b. 2 m
c. 20 M
d. 20 m
Question 3: What will be mole fraction of solute of a binary solution if that of its solvent is 0.2?
a. 0.8
b. 0.1
c. 0.7
d. 1.0
Question 4: Unit of molarity is
a. mol L^{-1}
b.mol kg^{-1}
c. mol^{-1} L
d. mol^{-1 }kg
Q.1
Q.2
Q.3
Q.4
b
a
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