Question

Derive the relation between molarity and molality. Calculate the molality of $1\mathrm{M}\mathrm{HCl}$ solution having a density $1.5365\text{g/mol}$

Answer 1

Part 1:

Molarity:

• It is denoted by M.
• Molarity (M) is defined as the number of moles of solute dissolved in per litre of solution.

$\text{Molarity =}\frac{\mathrm{moles}\mathrm{of}\mathrm{solute}}{\mathrm{litres}\mathrm{of}\mathrm{solution}}$

For example, the molecular mass of sulphuric acid(${\mathrm{H}}_2{\mathrm{SO}}_4$)is $98$.

Now if $98$ grams of sulphuric acid are present in 1 litre of the solution then it is 1 molar solution of Sulphuric acid and therefore molarity will be 1.

Molality:

• It is denoted by m.
• Molality (m) is defined as the number of moles of solute dissolved in per kilogram of solvent.

$\mathrm{Molality}=\frac{\mathrm{moles}\mathrm{of}\mathrm{solute}}{\mathrm{Kilograms}\mathrm{of}\mathrm{solvent}}$

For example, the molecular mass of sulphuric acid is $98$.

If $98$grams of sulphuric acid are present in 1 Kg of water then it is 1 molal solution of Sulphuric acid in water and therefore molality will be 1.

Relation between molarity and molality:

let the mass of the given solute = W

The volume of solution = V

The molality of the solution = m

The molar mass of solute = M’

Molarity of the solution = M

Weight of solvent = W’

$$\begin{gathered} \text{Molarity(M) =}\frac{\text{W × 1000}}{\text{M' × V}}\text{(1)} \\ \text{Molality (m) =}\frac{\mathrm{W}\times 1000}{{\mathrm{M}}^{\text{'}}\times {\mathrm{W}}^{\text{'}}}\text{(2)} \\ \mathrm{Density}\left(\mathrm{d}\right)=\frac{\mathrm{Mass}}{\mathrm{Volume}}=\frac{\mathrm{M}}{\mathrm{V}} \\ \mathrm{Density}\left(\mathrm{d}\right)=\frac{\mathrm{W}+{\mathrm{W}}^{\text{'}}}{\mathrm{V}} \\ \mathrm{From}\mathrm{equation}\left(1\right) \\ \mathrm{V}=\frac{\mathrm{W}\times 1000}{\mathrm{M}\times {\mathrm{M}}^{\text{'}}} \\ \mathrm{From}\mathrm{equation}\left(2\right) \\ {\mathrm{W}}^{\text{'}}=\frac{\mathrm{W}\times 1000}{{\mathrm{M}}^{\text{'}}\times \mathrm{m}}\left(3\right) \\ \mathrm{Now}\mathrm{adding}\mathrm{W}\mathrm{on}\mathrm{both}\mathrm{sides}\mathrm{in}\mathrm{equatiuon}\left(3\right),\mathrm{we}\mathrm{get} \\ {\mathrm{W}}^{\text{'}}+\mathrm{W}=\mathrm{W}+\frac{(\mathrm{W}\times 1000)}{\mathrm{m}\times {\mathrm{M}}^{\text{'}}}\left(4\right) \\ \mathrm{Now},\mathrm{on}\mathrm{multiplying}\mathrm{by}1000\mathrm{in}\mathrm{numerator}\mathrm{and}\mathrm{in}\mathrm{denominator},\mathrm{we}\mathrm{get} \\ {\mathrm{W}}^{\text{'}}+\mathrm{W}=(\mathrm{W}\times 1000)\left(\frac{{\mathrm{mM}}^{\text{'}}+1000}{{\mathrm{mM}}^{\text{'}}\times 1000}\right) \\ =\left(\frac{\mathrm{W}\times 1000}{{\mathrm{M}}^{\text{'}}}\right)(\frac{{\mathrm{M}}^{\text{'}}}{1000}+\frac{1}{\mathrm{m}})\left(5\right) \\ \mathrm{On}\mathrm{dividing}: \\ \frac{\mathrm{W}+\mathrm{W}\text{'}}{\mathrm{V}}=\mathrm{M}(\frac{\mathrm{M}\text{'}}{1000}+\frac{1}{\mathrm{m}}) \\ \mathrm{d}=\mathrm{M}(\frac{\mathrm{M}\text{'}}{1000}+\frac{1}{\mathrm{m}}) \\ \frac{\mathrm{d}}{\mathrm{M}}=\frac{{\mathrm{M}}^{\text{'}}}{1000}+\frac{1}{\mathrm{m}} \\ \frac{1}{\mathrm{m}}=\frac{\mathrm{d}}{\mathrm{M}}+\frac{\mathrm{M}\text{'}}{10000} \end{gathered}$$

Part 2:

Given density = $1.5365\text{g/mol}$

Molarity = $1\mathrm{M}$

The molecular weight of solute ($\mathrm{HCl}$) = $36.46\text{g/mol}$

$\text{Molality =}\frac{\mathrm{Molarity}}{(\mathrm{density}\mathrm{of}\mathrm{the}\mathrm{solution}-\mathrm{molarity})\times \mathrm{molecular}\mathrm{weight}\mathrm{of}\mathrm{solute}}$

$\mathrm{Molality}=\frac{1}{(1.5365-1)\times 36.46}$

$\text{Molality =}\frac{1}{19.56}$

$\mathrm{Molality}\left(\mathrm{m}\right)=0.05112\mathrm{m}$

Hence the molality of $1\mathrm{M}\mathrm{HCl}$ solution has a density $1.5365\text{g/mol = 0.05112 m}$