We always discuss a solution being diluted or concentrated; this is a qualitative way of expressing the concentration of the solution. A dilute solution means the quantity of solute is relatively very small, and a concentrated solution implies that the solution has a large amount of solute. But these are relative terms and do not give us the quantitative concentration of the solution.
So, to quantitatively describe the concentrations of various solutions around us, we commonly express levels in the following way, let’s study each method and determine the formulas for this method:
It is the amount of solute in grams present in 100 grams of the solution.
Therefore, the formula will be:
The ratio mass of solute to the mass of the solvent is the mass fraction. Thus, the 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., we have dissolved 10 grams of sugar in 90 grams of water.
It is the volume of solute in mL present in 100 mL solution.
The formula will be:
Volume Percentage = Volume of solute/Volume of Solution × 100
10% solution of HCl by volume means that 10 mL of liquid HCl is present in 100 mL of the solution.
It is the mass of solute present in 100 mL of solution. We can calculate the mass of the solute using the volume percentage.
The formula would be:
Mass by Volume Percentage = Mass of Solute/Volume of Solution × 100
It changes on changing temperature.
Example: A 10% (w/v) urea solution. = 10 gm of urea is present in 100 mL of solution.
The density of the solution is required to calculate the weight of solute and the weight of solvent.
The molarity of a solution gives the number of gram molecules of the solute present in one litre of the solution.
Molarity(M) = number of moles of solute/Volume of Solution in L
w = Mass of solute in gramsM = molecular weight of solute in gm/mol.
V = volume of solution in ml.
1 mol L^{1} solution of KCl means that 1 mol of KCl is dissolved in 1 L of water.
Unit of molarity: mol L^{1}
↠ Molarity is the most common way of representing the concentration of the solution.
↠ Molarity is dependent on temperature as M ∝ 1/T
↠ When a solution is diluted (x times), its molarity also decreases (by x times)
Question 1: Calculate the molarity of a solution containing 5 g of NaOH in 450 mL solution.
Solution: Moles of NaOH = 5g/40 g mol^{1} = 0.125 mol
Volume of the solution in litres = 450 mL / 1000 mL L^{1 }
Using equation (2.8),
= 0.278 mol L^{–1}
= 0.278 mol dm^{–3}
Molality of a solution is the number of moles of solute dissolved in 1 Kg of the solvent.
Molality (m) = Number of moles of solute/Mass of Solvent in Kg
w = mass of solute in grams
M = molecular wt of solute
W = mass of solvent in gram.
Thus, if one gram molecule of a solute is present in 1 kg of the solvent, the concentration of solutions is said to be one molal.
The unit of molarity is mol kg^{1}.
Molality is the most convenient method to express the concentration of solutions because it involves the mass of liquids rather than their volumes. It is also independent of the variation in temperature.
Question 2: Calculate the molality of 2.5 g of ethanoic acid (CH_{3}COOH) in 75 g of benzene.
Solution: Molar mass of C_{2}H_{4}O_{2} : 12 × 2 + 1 × 4 + 16 × 2 = 60 g mol^{–1}
Moles of C_{2}H_{4}O_{2} = 2.5/60 g mol^{1} = 0.0417 mol
Mass of benzene in kg = 75 g/1000 g kg^{–1} = 75 × 10^{–3} kg
Molality of C_{2}H_{4}O_{2}
The normality of a solution gives the number of gram equivalents of the solute present in one litre of the solution.
Normality (N) = Number of gram equivalent of solute/Volume of Solution in L
Equivalent mass = Molar mass/n  factor
No. of equivalent = Mass of solute/equivalent mass
= No. of moles of solute x n  factor
w = mass of solute in gram
V = volume of solution in ml
E = equivalent wt of solute
Thus, if one gram equivalent of a solute is present in one litre of the solution, the concentration of solutions is said to be 1 normal.
↠ 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
Relationship between normality and other Concentration Terms
↠ Molarity × Molecular mass= Strength of the solution (g/L)
↠ Normality × Equivalent mass = Strength of the solution (g/L)
↠ Molarity × Molecular mass= Normality x Equivalent mass
↠ Normality = n × Molarity
Mole fraction can be defined as the ratio of a number of moles of the component in the solution to the total number of moles of all components in the solution.
unity.
Mathematically, x_{1} + x_{2} + x_{3} …. + x_{i} =1
Question 3: Calculate the mole fraction of ethylene glycol (C_{2}H_{6}O_{2}) in a solution containing 20% of C_{2}H_{6}O_{2} by mass.
Solution: Assume that we have 100 g of solution (one can start with any amount of solution because the results obtained will be the same). The solution will contain 20 g of ethylene glycol and 80 g of water.
Molar mass of C_{2}H_{6}O_{2 }= 12 × 2 + 1 × 6 + 16 × 2 = 62 g mol^{–1}
Mole fraction of water can also be calculated as: 1 – 0.068 = 0.932.
Question 4: Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Solution.
Total mass of the solution = 100 g
Mass of benzene = 30 g.
∴ Mass of carbon tetrachloride = (100  30)g = 70 g
Molar mass of benzene (C_{6}H_{6}) = (6 × 12 + 6 × 1) g mol^{– 1} = 78 g mol^{1}
∴ Number of moles of C_{6}H_{6} =30/78 mol
= 0.3846 mol
Molar mass of carbon tetrachloride (CCl_{4}) = 1 × 12 + 4 × 355 = 154 g mol^{1}
∴ Number of moles of CCl_{4} = 70/154 mol = 0.4545 mol
Mole fraction of C_{6}H_{6}
Question 5: A tank is charged with a mixture of 1.0 × 10^{3} mol of oxygen and 4.5 × 10^{3} mol of helium. Calculate the mole fraction of each gas in the mixture.
Solution.
The given parameters are
N_{He} = 4.5 × 10^{3} mol and N_{O}_{2} = 1.0 × 10^{3} mol
Mole fraction can be calculated as
X_{He} = 4.5 × 10^{3} mol / 4.5 × 10^{3} mol + 1.0 × 10^{3} mol
X_{He }= 4.5 mol / 5.5 mol
X_{He }= 0.82
X_{O2} = 2.0 × 10^{3} mol / 4.5 × 10^{3} mol + 1.0 × 10^{3} mol
X_{O2 }= 1.0 × 10^{3} / 5.5 × 10^{3}
X_{O2 }= 0.18
Question 6: Determine the mole fraction of methanol CH_{3}OH and water in a solution prepared by dissolving 4.5 g of alcohol in 40 g of H_{2}O. M of H_{2}O is 18 and M of CH_{3}OH is 32.
Solution.
Moles of CH_{3}OH = 4.5 / 32 = 0.14 mole
Moles of H_{2}O = 40 / 18 = 2.2 moles
Therefore, according to the equation
mole fraction of CH_{3}OH = 0.14 / 2.2 + 0.14
= 0.061
Question 7: What is the mole fraction of carbon tetrachloride (CCl4) in solution if 3.47 moles of CCl4 is dissolved in 8.54 moles of benzene (C_{6}H_{6})?
Solution.
Question 8: What is the mole fraction of formaldehyde (CH_{2}O) in solution if 25.7 grams of CH_{2}O is dissolved in 3.25 moles of carbon tetrachloride (CCl_{4})?
Solution.
The grams of acetone will need to be converted into moles in order to solve for mole fraction.
Question 9: 0.100 mole of NaCl is dissolved into 100.0 grams of pure H_{2}O. What is the mole fraction of NaCl?
Solution.
100.0 g / 18.0 g mol^{}^{1} = 5.56 mol of H_{2}O
Add that to the 0.100 mol of NaCl = 5.56 + 0.100 = 5.66 mol total
Mole fraction of NaCl = 0.100 mol / 5.66 mol = 0.018
The mole fraction of the H_{2}O:
5.56 mol / 5.66 mol = 0.982
Question 10: A solution is prepared by mixing 25.0 g of water, H_{2}O, and 25.0 g of ethanol, C_{2}H_{5}OH. Determine the mole fractions of each substance.
Solution.
(1) Determine the moles of each substance:
H2O ⇒ 25.0 g / 18.0 g/mol = 1.34 mol
C_{2}H_{5}OH ⇒ 25.0 g / 46.07 g/mol = 0.543 mol
(2) Determine mole fractions:
H_{2}O ⇒ 1.34 mol / (1.34 mol + 0.543 mol) = 0.71
C_{2}H_{5}OH ⇒ 0.543 mol / (1.34 mol + 0.543 mol) = 0.29
Question 11: A solution contains 10.0 g pentane, 10.0 g hexane and 10.0 g benzene. What is the mole fraction of hexane?
Solution.
(1) You need to determine the moles of pentane, hexane and benzene:
to do this, you need the molecular weights. Here are the formulas:
pentane: C_{5}H_{12}
hexane: C_{6}H_{14}
benzene: C_{6}H_{6}
(2) When you have the moles of each, add them together.
(3) Then, divide the moles of hexane by the total.
Question 12: The molality of an aqueous solution of sugar (C_{12}H_{22}O_{11}) is 1.62m. Calculate the mole fractions of sugar and water.
Solution.
(1) Molality is moles solute / kg of solvent. Therefore we know our solution is:
1.62 mol C_{12}H_{22}O_{11}
1.00 kg = 1000 g of water
(2) Calculate the moles of water present:
1000 g / 18.0152 g/mol = 55.50868 mol
(3) Determine the mole fraction of the sugar:
1.62 mol / (1.62 mol + 55.50868 mol) = 0.028357 = 0.0284 (to three sf)
(4) You can calculate the mole fraction of the water by subtraction.
Question 13: How many grams of water must be used to dissolve 100.0 grams of sucrose (C_{12}H_{22}O_{11}) to prepare a 0.020 mole fraction of sucrose in the solution?
Solution.
(1) Determine moles of sucrose:
100.0 g / 342.2948 g/mol = 0.292145835 mol
(2) Determine moles of water required to make the solution 0.020 mole fraction of sucrose:
0.020 = 0.292 / (0.292 + x)
(0.020) (0.292 + x) = 0.292
0.00584 + 0.02x = 0.292
0.02x = 0.28616
x = 14.308 mol of H_{2}O
Comment: you can also do this:
0.292 is to 0.02 as x is to 0.98
(3) Determine grams of water:
14.308 mol × 18.015 g/mol = 258.0 g
Question 14: Surprisingly, water (in the form of ice) is slightly soluble in liquid nitrogen. At 196 °C, (the boiling point of liquid nitrogen) the mole fraction of water in a saturated solution is 1.00 × 10^{5}. Compute the mass of water that can dissolve in 1.00 kg of boiling liquid nitrogen.
Solution.
(1) Use the definition of mole fraction to set up the following:
χ_{water} = moles water / (moles water + moles nitrogen)
1.00 × 10^{5} = x / (x + 71.3944041)
I'm going to carry some guard digits until the end of the calculation.
(2) Some algebra:
(1.00 x 10^{5}) (x) + 7.139440411 × 10^{4} = x
0.99999x = 7.139440411 x 10^{4}
x = 7.139511806 × 10^{4} mol of H_{2}O
(3) Calculate grams of water from moles of water:
7.139511806 × 10^{4} mol × 18.0152 g/mol = 1.2862 × 10^{2 }g
1.29 × 10^{2} g (to three sf)
Question 15: What is the mole fraction of cinnamic acid in a mixture that is 50.0% weight urea in cinnamic acid (urea = 60.06 g/mol; cinnamic acid = 148.16 g/mol)
Solution.
(1) Let us assume 100.0 g of this mixture are present.
Therefore:
50.0 g is urea
50.0 g is cinnamic
(2) Convert grams to moles:
urea: 50.0 g / 60.06 g/mol = 0.8325 mol
cinnamic acid: 50.0 g / 148.16 g/mol = 0.3375 mol
(3) Determine mole fraction of cinnamic acid:
0.3375 mol / 1.1700 mol = 0.2885
When a solute is present in trace quantities, it is convenient to express the concentration of solutions in parts per million (ppm). The formula is as follows:
In case of mass, we may express it as:
In case of volume, we may express it as:
So, we can express the concentration of solutions in parts per million as mass to mass, volume to volume and mass to volume form.
It is the number of mass in grams present per litre of solution. In case, formula mass is equal to molecular mass, formality is equal to molarity.
Where,
It is defined as the weight of solute per litre (1000 mL) of solution.
Example: 10% w/v sucrose solution, then specify its concentration in g/L.
In 100 mL solution, weight of sucrose = 10 g
Therefore, in 1000mL solution= 10/100 = 100 g
Hence strength will be 100g/L
These solutions are mixed; molarity of mixed solution may be given as:
where, MR is the Resultant molarity
V_{1 }+ V_{2} + V_{3} = Resultant volume after mixing
Note: Molarity is dependent on volume; therefore, it depends on temperature.
1M  molar solution, i.e, molarity is l 
0.5 M or M/2  Semimolar 
0.1 M or M/2  Decimolar 
0.01 M or M/100  Centimolar 
0.001 M or M/100  Millimolar 
The dilution formula is giving by the following expression:
M_{1}V_{1} = M_{2}V_{2} 
Problem 1. Calculate the masses of cane sugar and water required to prepare 250 g of 25% cane sugar solution.
Solution.
Mass percentage of cane sugar = 25
We know that,
Mass percentage = Mass of solute/Mass of solution × 100
25 = Mass of cane sugar/250g × 100
Mass of cane sugar = 25 ×250/100 = 62.5 g
Mass of water (250 62.5)g = 187.5 g
Problem 2. A tank is charged with a mixture of 1.0 x 10^{3} mol of oxygen and 4.5 x 10^{3} mol of helium. Calculate the mole fraction of each gas in the mixture.
Solution.
The given parameters are
n_{He} = 4.5 x 10^{3} mol and nO_{2} = 1.0 x 10^{3} mol
Mole fraction can be calculated as
x_{He} = 4.5 x 10^{3} mol / (4.5 x 10^{3} mol + 1.0 x 10^{3} mol)
=4.5 mol / 5.5 mol = 0.82
xO_{2} = 2.0 x 10^{3} mol / (4.5 x 10^{3} mol + 1.0 x 10^{3} mol)
= 1.0 x 10^{3} / 5.5 x 10^{3}
= 0.18
Problem 3. The density of a solution containing 13% by mass of sulphuric acid is 1.09 g/mL. Calculate the molarity and normality of the solution.
Solution.
Volume of 100 g of the solution
= 100/d = 100/1.09 mL
= 100/1.09 × 1000 litre
= 1/1.09 × 10 litre
Number of moles of H_{2}so_{4} in 100 g of the solution = 13/98
Molarity = No. of moles of H_{2}So_{4}/Volume of Solu. in litre = 13/98 × 1.09 × 10/1
= 1.445 M
We know that,
Normality = Molarity × n
Problem 4. What is the mole fraction of carbon tetrachloride (CCl_{4}) in solution if 3.47 moles of CCl_{4} is dissolved in 8.54 moles of benzene (C_{6}H_{6})?
Solution.
Mole Fraction CCI_{4} (X_{CCI}_{4}) = moles of CCI_{4}/total moles
X_{CCI}_{4} = 3.47 moles CCI_{4}/3.47 moles CCI_{4} + 8.54 moles C_{6}H_{6}_{ }
XCCI4 = 0.289
Problem 5. What is the mole fraction of formaldehyde (CH_{2}O) in solution if 25.7 grams of CH_{2}O is dissolved in 3.25 moles of carbon tetrachloride (CCl_{4})?
Solution.
The grams of acetone will need to be converted into moles in order to solve for mole fraction.
Mole Fraction CH_{2}O(X_{CH}_{2}_{O}) = moles of CH_{2}O/total moles
X_{CH}_{2}_{O = 0.856 moles CH}_{2}_{O/0.856 moles CH}_{2}_{O + 3.25 moles CCI}_{4}
X_{CH}_{2}_{O = 0.208}
Problem 6. 0.100 mole of NaCl is dissolved into 100.0 grams of pure H_{2}O. What is the mole fraction of NaCl?
Solution.
100.0 g / 18.0 g mol^{1} = 5.56 mol of H_{2}O
Add that to the 0.100 mol of NaCl = 5.56 + 0.100 = 5.66
mol total Mole fraction of NaCl = 0.100 mol / 5.66 mol = 0.018
The mole fraction of the H_{2}O:
5.56 mol / 5.66 mol = 0.982
Problem 7. A solution is prepared by mixing 25.0 g of water, H_{2}O, and 25.0 g of ethanol, C_{2}H_{5}OH. Determine the mole fractions of each substance.
Solution.
(1) Determine the moles of each substance:
H_{2}O ⇒ 25.0 g / 18.0 g/mol = 1.34 mol
C_{2}H_{5}OH ⇒ 25.0 g / 46.07 g/mol = 0.543 mol
(2) Determine mole fractions:
H_{2}O ⇒ 1.34 mol / (1.34 mol + 0.543 mol) = 0.71
C_{2}H_{5}OH ⇒ 0.543 mol / (1.34 mol + 0.543 mol) = 0.29
Problem 8. A solution contains 10.0 g pentane, 10.0 g hexane and 10.0 g benzene. What is the mole fraction of hexane?
Solution.
Try it yourself Using hints given below !
(1) You need to determine the moles of pentane, hexane and benzene: to do this, you need the molecular weights. Here are the formulas:
pentane: C_{5}H_{12}
hexane: C_{6}H_{14}
benzene: C_{6}H_{6}
(2) When you have the moles of each, add them together.
(3) Then, divide the moles of hexane by the total.
Problem 9. The molality of an aqueous solution of sugar (C_{12}H_{22}O_{11}) is 1.62m. Calculate the mole fractions of sugar and water.
Solution.
(1) Molality is moles solute / kg of solvent. Therefore we know our solution is:
1.62 mol C_{12}H_{22}O_{11}
1.00 kg = 1000 g of water
(2) Calculate the moles of water present:
1000 g / 18.0152 g/mol = 55.50868 mol
(3) Determine the mole fraction of the sugar:
1.62 mol / (1.62 mol + 55.50868 mol) = 0.028357 = 0.0284 (to three sf)
(4) You can calculate the mole fraction of the water by subtraction.
Problem 10. How many grams of water must be used to dissolve 100.0 grams of sucrose (C_{12}H_{22}O_{11}) to prepare a 0.020 mole fraction of sucrose in the solution?
Solution.
(1) Determine moles of sucrose:
100.0 g / 342.2948 g/mol = 0.292145835 mol
(2) Determine moles of water required to make the solution 0.020 mole fraction of sucrose:
0.020 = 0.292 / (0.292 + x)
(0.020) (0.292 + x) = 0.292
0.00584 + 0.02x = 0.292
0.02x = 0.28616
x = 14.308 mol of H_{2}O
Comment: you can also do this:
0.292 is to 0.02 as x is to 0.98
(3) Determine grams of water:
14.308 mol x 18.015 g/mol = 258.0 g
Problem 11. Surprisingly, water (in the form of ice) is slightly soluble in liquid nitrogen. At 196 °C, (the boiling point of liquid nitrogen) the mole fraction of water in a saturated solution is 1.00 x 10^{5}. Compute the mass of water that can dissolve in 1.00 kg of boiling liquid nitrogen.
Solution.
(1) Use the definition of mole fraction to set up the following:
χwater = moles of water / (moles of water + moles of nitrogen)
1.00 x 10^{5} = x / (x + 71.3944041)
I'm going to carry some guard digits until the end of the calculation.
(2) Some algebra: (1.00 x 10^{5}) (x) + 7.139440411 x 10^{4} = x
0.99999x = 7.139440411 x 10^{4}
x = 7.139511806 x 10^{4} mol of H_{2}O
(3) Calculate grams of water from moles of water:
7.139511806 x 10^{4} mol x 18.0152 g/mol = 1.2862 x 10^{2} g
1.29 x 10^{2} g (to three sf)
Problem 12. What is the mole fraction of cinnamic acid in a mixture that is 50.0% weight urea in cinnamic acid (urea = 60.06 g/mol; cinnamic acid = 148.16 g/mol)
Solution.
(1) Let us assume 100.0 g of this mixture are present. Therefore:
50.0 g is urea
50.0 g is cinnamic
(2) Convert grams to moles: urea:
50.0 g / 60.06 g/mol = 0.8325 mol
cinnamic acid: 50.0 g / 148.16 g/mol = 0.3375 mol
(3) Determine mole fraction of cinnamic acid:
0.3375 mol / 1.1700 mol = 0.2885
Problem 13. 5 g of NaCl is dissolved in 1000 g of water. If the density of the resulting solution is 0.997 g per cc, calculate molality, molarity, normality and mole fraction of the solute.
Solution.
Mole of NaCl = 5/58.5 = 0.0854 (Mol. wt. of NaCl = 58.5)
Molality = Moles/Wt. of solvent in gram x 1000
= 0.0854/1000 x 1000 = 0.0854 m
Volume of the solution= Wt. of solution in gram/Density in gram / cc
Again by definition
Molarity = Moles/Volume of solution in litre
= 0.0854/1.008
= 0.085 M
∴ Normality = 0.085 N (for NaCl, eq. wt. = mol. wt)
Further, Mole of H_{2}O = 1000/18 = 55.55
(1000 gram of water = 1000 ml of water, because density = 1 g/cc)
Total mole = Mole of NaCl + Mole of H_{2}O
= .0854 + 55.55 = 55.6354
Mole fraction of NaCl = 7.85/98.66 = 0.0854/55.6409 = 1.535 x 10^{–3}
Problem 14. If 20 ml of ethanol (density = 0.7893 gm / ml) is mixed with 40 ml of water (density = 0.9971 gm/ml) at 25°C, the final solution has density of 0.9571 gm / ml. Calculate the percentage change in total volume of mixing. Also calculate the molality of alcohol in the final solution.
Solution.
Mass of ethanol = v x d
W = 20 × 0.7893 gm = 15.786 gm
Mass of water = v × d
= 40 x 0.9971 gm = 39.884 gm
Total volume = 60 ml
Total mass = 15.786 + 39.884 = 55.67 gm
Let the volume of solution = x ml
X = mass/density = 55.67/0.9571 = 58.165 ml
Change in volume = 60 – 58.165 = 1.835 ml
% change in volume = 1.835/60 x 100
= 3.05%
Molality, m = w/m x 1000/W (in gm) = 15.786/46 x 1000/39.884
Problem 15. Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Solution.
Total mass of the solution = 100 g
Mass of benzene = 30 g.
∴ Mass of carbon tetrachloride = (100  30)g = 70 g
Molar mass of benzene (C_{6}H_{6}) = (6 × 12 + 6 × 1) g mol^{– 1} = 78 g mol^{1}
∴ Number of moles of C_{6}H_{6} =30/78 mol = 0.3846 mol
Molar mass of carbon tetrachloride (CCl_{4}) = 1 × 12 + 4 × 355 = 154 g mol^{1}
∴ Number of moles of CCl_{4} = 70/154 mol = 0.4545 mol
Mole fraction of C_{6}H_{6} =
Problem 16. Determine the mole fraction of methanol CH_{3}OH and water in a solution prepared by dissolving 4.5 g of alcohol in 40 g of H_{2}O. Molar mass of H_{2}O is 18 and Molar mass of CH_{3}OH is 32.
Solution.
Moles of CH_{3}OH = 4.5/32 = 0.14
mole Moles of H_{2}O = 40/18 = 2.2 moles
Therefore, according to the equation
mole fraction of CH_{3}OH = 0.14/2.2 + 0.14
= 0.061
135 videos348 docs182 tests

1. What is the mass/weight percentage or percentage by mass/weight expression of concentration of a solution? 
2. How is volume percentage used to express the concentration of a solution? 
3. What does parts per million (ppm) indicate in the expression of concentration of a solution? 
4. How is formality used as an expression of concentration of a solution? 
5. How is strength (g/L) used to express the concentration of a solution? 

Explore Courses for NEET exam
