<< Chapter < Page | Chapter >> Page > |
Molality (m) is also a measurement of molar concentration like molarity (M). We have seen that molarity(M) is a convenient measurement of the concentration of solution as it allows us to directly compute moles of solute present in the solution. There is, however, a problem in reporting concentration of a solution in terms of molarity. Recall that it is equal to numbers of moles of solute divided by volume of solution in litres. The molar ratio has volume of the solution in the denominator. This means that molarity(M) of a given solution will change with temperature as volume of solution will change with temperature. This is a major handicap as reported concentration needs to be reliable and constant anywhere irrespective of temperature. In this backdrop, molality(m) is measurement of concentration aiming to remove this shortcoming associated with measurement in molarity (M).
Molality(m) also differs to other measurements in yet another important aspect. It involves the ratio of measurement of solute and solvent – not that of solute and solution. This difference is important to be kept in mind while computing quantities and converting measurement units from one to another.
The major objective of this module is to develop skills to convert measurement of concentration from one measurement type to another.
Molality of a solution with respect to solute is defined as :
$$\text{Molality(m)}=\frac{\text{Moles of solute (B)}}{\text{Mass of solvent (A) in Kg}}$$
$$\Rightarrow m=\frac{{n}_{B}}{{w}_{\mathrm{Akg}}}$$
Its unit is moles/ kg. In case we consider mass of solvent in gm, then the expression of molality is given as :
$$\Rightarrow m=\frac{{n}_{B}}{{g}_{A}}X1000=\frac{\text{Milli-moles of B}}{{g}_{A}}$$
If mass of solvent of a solution of known molality is known, then number of moles of solute is obtained as :
$$\Rightarrow \text{Moles of solute, B}=\text{Molality}X\text{Mass of solvent in kg}$$
Similarly,
$$\Rightarrow \text{Milli-moles of B}=\text{Molality}X\text{Mass of solvent (A) in gm}$$
Problem : 11.7 gm of sodium chloride is dissolved in 400 ml of water. Find molality of the solution.
Solution : Here,
$$\Rightarrow \text{Moles of sodium chloride}=\frac{11.7}{\left(23+35.5\right)}=\frac{11.7}{58.5}=0.2$$
$$\Rightarrow \text{Mass of the solvent}=400X1=400\phantom{\rule{1em}{0ex}}gm=0.4\phantom{\rule{1em}{0ex}}kg$$
$$\Rightarrow \text{Molality, m}=\frac{0.2}{0.4}=0.5\phantom{\rule{1em}{0ex}}m$$
Molality and Molarity are linked to each other through density of solution. Beginning with the definition of molarity, a solution of molarity “M” means that 1 litre of solution contains “M” moles of solute. If the density of the solution is “d” in gm/cc, then
$$\text{Mass of 1 litre solution in gm}=1000d$$
$$\text{Mass of the solute in gm in 1 litre solution}=\text{nos of moles}X\text{molecular weight}=M{M}_{O}$$
$$\Rightarrow \text{Mass of the solvent in gm in 1 litre solution}=1000d-M{M}_{O}$$
We need to calculate mass of solvent in kg to calculate molality(m) :
$$\Rightarrow \text{Mass of the solvent in kg in 1 litre solution}=\left(1000d-M{M}_{O}\right)/1000$$
Hence, molality,
$$m=\frac{{n}_{B}}{{W}_{\mathrm{Akg}}}=\frac{M}{\frac{1000d-M{M}_{O}}{1000}}=\frac{1000M}{1000d-M{M}_{O}}$$
We should note that "density of solution (d)" and "strength of solution (S)" differ. Density of solution (d) is ratio of mass of solution (solute + solvent) in gm and volume of solution in cc. It has the unit of gm/cc. On the other hand, strength of solution (S) is ratio of mass of solute in gm and volume of solution in litre. It has the unit of gm/litre.
Problem : The density of 3M sodium thiosulphate (Na2S2O3) solution is 1.25 gm/cc. Calculate molality of ${\mathrm{Na}}^{+}$ and $${S}_{2}{O}_{{3}^{--}}$$ ions.
Solution : We can use the formula to calculate molality of the sodium thiosulphate :
$$\Rightarrow m=\frac{1000M}{1000d-M{M}_{O}}=\frac{1000X3}{1000X1.25-3X\left(2X23+2X32+3X16\right)}$$
$$\Rightarrow m=\frac{3000}{1250-474}=\frac{3000}{776}=3.866$$
Alternatively , we can proceed with the basic consideration in place of using formula Since molarity of solution is 3M, it means that 1 litre of solution contains 3 moles of sodium thiosulphate. We can use density to find the mass of the 1 litre solvent.
$$\Rightarrow \text{Mass of solution}=1000X1.25=1250\phantom{\rule{1em}{0ex}}gm$$
$$\Rightarrow \text{Mass of 3 moles of sodium thiosulphate}=3{M}_{N{a}_{2}{S}_{2}{O}_{3}}=3X\left(2X23+2X32+3X16\right)$$
$$\Rightarrow \text{Mass of 3 moles of sodium thiosulphate}=3X\left(46+64+48\right)=3X158=474\phantom{\rule{1em}{0ex}}gm$$
$$\Rightarrow \text{Mass of solvent}=1250-474=776\phantom{\rule{1em}{0ex}}gm=0.776\phantom{\rule{1em}{0ex}}kg$$
$$\Rightarrow \text{Molality of sodium thiosulphate}=\frac{3}{0.776}=3.866\phantom{\rule{1em}{0ex}}m$$
We are, however, required to calculate molality of ions. We see that one mole is equivalent to 2 moles of sodium ion and 1 mole of thiosulphate ion :
$${\mathrm{Na}}_{2}{S}_{2}{O}_{3}=2N{a}^{+}+{S}_{2}{O}_{{3}^{--}}$$
Hence,
$$\Rightarrow \text{Molality of}{\mathrm{Na}}^{+}=2X3.866=7.732\phantom{\rule{1em}{0ex}}m$$
$$\Rightarrow \text{Molality of}{S}_{2}{O}_{{3}^{--}}=3.866\phantom{\rule{1em}{0ex}}m$$
Working on the relation of molality developed in previous section :
$$m=\frac{1000M}{1000d-M{M}_{O}}$$
$$\Rightarrow \frac{1}{m}=\frac{1000d-M{M}_{O}}{1000M}=\frac{d}{M}-\frac{{M}_{O}}{1000}$$
$$\Rightarrow d=M\left(\frac{1}{m}+\frac{{M}_{O}}{1000}\right)$$
Problem : The molality and molarity of a solution of sulphuric acid are 90 and 10 respectively Determine density of the solution.
Solution : Using relation :
$$\Rightarrow d=M\left(\frac{1}{m}+\frac{{M}_{O}}{1000}\right)$$
$$\Rightarrow d=10\left(\frac{1}{90}+\frac{98}{1000}\right)=10X\left(0.011+0.98\right)=10X0.991=9.91\phantom{\rule{1em}{0ex}}gm/cc$$
We need to know the moles of solute and mass of solvent in kg to determine molality. Now, strength of solution (S) is equal to mass of the solute in gm in 1 litre of solution. Hence,
$$\text{Mass of the solute in gm in 1 litre of solution}=S$$
$$\text{Moles of the solute}=\frac{S}{{M}_{O}}$$
$$\text{Mass of 1 litre solution in gm}=1000d$$
$$\text{Mass of the solvent in gm in 1 litre of solution}=1000d-S$$
$$\text{Mass of the solvent in kg in 1 litre of solution}=\frac{\left(1000d-S\right)}{1000}$$
The molality is :
$$\Rightarrow m=\frac{{n}_{B}}{{W}_{\mathrm{Akg}}}=\frac{\frac{S}{{M}_{O}}}{\frac{\left(1000d-S\right)}{1000}}=\frac{1000S}{{M}_{O}\left(1000d-S\right)}$$
Problem : A solution has 392 gm of sulphuric acid per litre of solution. If the density of the solution is 1.25 gm/cc, find molality of the solution.
Solution : Using relation :
$$\Rightarrow m=\frac{1000S}{{M}_{O}\left(1000d-S\right)}=\frac{1000X392}{98X\left(1000X1.25-392\right)}=4.73\phantom{\rule{1em}{0ex}}m$$
Molality is defined as :
$$m=\frac{{n}_{B}}{{W}_{Akg}}=\frac{{n}_{B}}{{g}_{A}}X1000=\frac{{n}_{B}}{{n}_{A}{M}_{A}}X1000$$
$$\Rightarrow {n}_{B}=\frac{m{n}_{A}{M}_{A}}{1000}$$
On the other hand, mole fraction with respect to solute B is given by :
$${\chi}_{B}=\frac{{n}_{B}}{{n}_{A}+{n}_{B}}$$
Substituting for nA, we have :
$$\Rightarrow {\chi}_{B}=\frac{\frac{m{n}_{A}{M}_{A}}{1000}}{{n}_{A}+\frac{m{n}_{A}{M}_{A}}{1000}}=\frac{m{M}_{A}}{1000+m{M}_{A}}$$
$$\Rightarrow 1000{\chi}_{B}+m{M}_{A}{\chi}_{B}=m{M}_{A}$$
$$\Rightarrow m{M}_{A}\left(1-{\chi}_{B}\right)=1000{\chi}_{B}$$
$$\Rightarrow m=\frac{1000{\chi}_{B}}{\left(1-{\chi}_{B}\right){M}_{A}}$$
Similarly, we can express molality in terms of mole fraction with respect to solvent (A) as :
$$\Rightarrow m=\frac{1000\left(1-{\chi}_{A}\right)}{{\chi}_{A}{M}_{A}}$$
Notification Switch
Would you like to follow the 'Stoichiometry' conversation and receive update notifications?