Molarity

It is defined as the number of moles of solute per litre (or per dm³) of the solution. It is denoted by M. It is also known as molar concentration.

$${Molarity\ =\ \frac{No.\ of\ moles\ of\ solute}{Volume\ of\ solution\ in\ litre}}$$

$${M\ =\ \frac{n_{B}}{V\ (in\ l)}}$$
where, $n_{B}$ is the number of moles of solute.
(i) Its SI unit is $molL^{-1}$ or $M$.
(ii) Solution in terms of molarity generally expressed as: $1\ M$ = Molar solution; $2\ M$= Molarity of 2; $\frac{M}{2}$ = Semimolar solution; $\frac{M}{10}$ = Decimolar solution; $\frac{M}{100}$ = Centimolar solution; $\frac{M}{1000}$ = Millimolar solution, etc.
(iii) Molarity depends on temperature as, $\ $ $M\propto \frac{1}{T}$ i.e., with rise in temperature of a solution, its volume increases and hence its molarity decreases.
(iv) When a solution is diluted $x$ times, its molarity also decreases by $x$ times.

Numerical based on different formulas of molarity

(i) $M\ =\ \frac{n_{B}}{V\ (in\ l)}$,$\ \ \ \ $ (ii)$M\ =\ \frac{n_{B}\ \times\ 1000}{V\ (in\ ml)}$ $\ \ \ \ $(iii)$M\ =\ \frac{w_{B}\ \times\ 1000}{M_{B}\ \times\ V\ (in\ ml)}$ (where, $w_{B}$ = weight of the solute in gram, $M_{B}$ = Molar mass of the solute.)

Additional formula used: (a) Volume of solution =$\frac{mass\ of \ solution}{density\ of\ solution}$; (b) a % by mass means a g of solute per 100 g of solution.

(1) Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in enough water to form 250 mL of the solution. (Na = 23, O = 16, H = 1)
(2) Calculate the molarity of sodium carbonate (Na₂CO₃) solution, 5.3 g of which have been dissolved per 250 mL of the given solution. (Na = 23, C = 12, O = 16)
(3) Calculate the molarity of 0.365 g of pure HCl gas dissolved in 50 mL of solution.
(4) Calculate the molarity of commercially available concentrated solution which contains 38% HCl by mass. Given, density of solution = 1.19 g mL^-1.
(5) Calculate the molarity of 1 kg of water. Given the density of water = 1 kg/L.
(6) In 1200 g solution, 12 g urea is present. If the density of the solution is 1.2 g/mL, then calculate the molarity of the solution.
(7) Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g mL^-1. Calculate the molarity of a solution.

(iv) $No.\ of \ millimoles\ =\ \frac{w_{B}\ (in\ gram)}{M_B\ (in\ g\ mol^{-1})}\ \times\ 1000$,$\ \ \ \ $; $\ \ \ \ $ (v) $Molarity =\ \frac{no.\ of\ millimoles}{Volume\ of \ solution\ (in\ mL)}$

(1) How many millimoles are present in 0.56 g of KOH?
(2) How many millimoles are present in 120 g of glucose?
(3) Determine the molarity of a solution containing 3 millimoles of solute in 30 mL of solution.
(4) Calculate the molarity of glucose, 24 g of which is dissolved in enough water to form 750 mL of solution.

(vi) $Molarity\ =\ \frac{Strength\ of\ solution\ (gL^{-1})}{Molar\ mass\ of\ solute}$$\ \ \ \ $ (vii) $wt.\ of\ solute\ (w_{B}) = \frac{Molarity (M)\ \times\ Mol. wt.(M_{B})\ \times\ Volume\ of\ solution(V)\ in\ mL}{1000}$$\ \ \ \ $

(viii) $percent\ wt.\ of\ solute = \frac{wt.\ of\ solute}{wt.\ of\ solution} \ \times\ 100$

(1) Calculate the molecular weight of NaOH if its 0.5 M solution contains 20 g NaOH per litre.
(2) The density of a 3M sodium thiosulphate (Na₂S₂O₃) solution is 1.25 g/mL. Calculate the percent by weight of sodium thiosulphate. (Na = 23, S = 32, O = 16)
(3) Calculate the weight of oxalic acid (H₂C₂O₄.2H₂O) that will be required to prepare a 1.5 L (M/20) solution.
(4) Calculate the mass of sodium acetate (CH₃COONa) required to make 500 mL of 0.375 molar aqueous solution.

(ix) $Molarity\ =\ \frac{10\ \times\ sp.\ gravity\ of\ the\ solution\ \times\ wt\%\ of\ solute}{Mol.\ wt.\ of\ the\ solute}$, where specific gravity means density of solution.

(1) The density of 10% by mass of KCl solution in water is 1.06 g/mL. Calculate the molarity of the solution.
(2) Calculate the concentration of nitric acid (HNO₃) in moles per litre in a sample that has a density, 1.41 g/mL and the mass percent of nitric acid in it being 69%.
(3) Concentrated sulphuric acid has a density of 1.9 g/mL and is 99% H₂SO₄ by weight. Calculate the molarity of H₂SO₄ in this acid.
(4) Commercially available concentrated hydrochloric acid contains 38% HCl by mass and has density 1.19 g/cm³. Calculate the molarity of this solution.
(5) Calculate the molarity of the phosphoric acid solution of sp. gravity 1.426 and containing 60% by weight of pure H₃PO₄. (P = 31)
(6) Calculate the molarity of a solution of H₂SO₄ (sp. gr. 1.98) containing 27% H₂SO₄ by mass.

(x) In case of dilution, (a) $M_{1}V_{1}\ =\ M_{2}V_{2}$, where, V₂ = V₁ + V and V = volume of water added for dilution. (b) $V = \left( {\frac{{{M_1} – {M_2}}}{{{M_1}}}} \right) \times {V_1}$, where, V₁ = initial volume of solution.

(1) Calculate the molarity of a solution obtained by mixing 250 mL of 0.5 M NaOH and 150 mL of water.
(2) If the density of methanol is 0.793 kg L^-1, what is its volume needed for making 2.5 L of its 0.25 M solution?
(3) What volume of water should be added to 400 mL of 0.5 M HCl so as to get a 0.2 M solution?

(xi) Mixing of two solutions of different volumes and molarities, if M = resultant molarity of mixture and V = resultant volume of mixture then,
$$M = \left( {\frac{{{M_1}{V_1} + {M_2}{V_2}}}{V}} \right)$$
where, (a) V = V₁ + V₂, if there is no chemical reaction in the mixture, (b) V ≠ V₁ + V₂, if there takes place chemical reaction in mixture.

(1) Calculate the molarity of a solution obtained by mixing 300 mL of 0.5 M NaOH and 200 mL of 0.2 M KOH solution.

Problems

(1) Equal volumes of two solutions contain 50 grams sodium chloride and 50 grams potassium chloride, respectively. Are their molarities equal?
(2) How many millimoles of concentrated sulphuric acid of density 1.84 g/mL containing 98% H₂SO₄ by mass are required to make 200 mL of 0.25 M solution.
(3) Calculate the volume of a 0.1 M HCl that are needed to react completely with 1 g of a mixture of NaHCO₃ and Na₂CO₃ containing equimolar amounts of both.
(4) A solution contains 410.3 g of H₂SO₄ per litre of solution at 20⁰ C. If the density is 1.243 g/mL, what will be its molarity?
(5) Under heating a litre of $\frac{M}{2}$ HCl solution, 2.675 g hydrogen chloride is lost due to evaporation and the volume of the solution shrinks to 750 ml. Calculate (a) the molarity of the resulting solution. (b) the number of millimoles of HCl in 100 ml of the final solution.
(6) How many grams of concentrated nitric acid solution be used to prepare 250 mL of 2.0 M HNO₃? The concentrated acid is 70%.
(7) 6.02 x 10²⁰ molecules of urea are present in 100 mL of its solution. Calculate the molarity of solution. (Urea = NH₂CONH₂)
(8) HCl is 80% ionised in 0.01M aqueous solution. What is its equilibrium molarity? What are other species present? What are their molar concentrations?
(9) How many mL of a 0.045 M BaCl₂ contain 15.0 g BaCl₂ (208 g mol^-1)?
(10) The concentration of cholesterol (C₂₇H₄₆O) in normal blood is approximately 0.005 M. How many grams of cholesterol are in 750 mL of blood?