The molality of a given solution (m) is defined as the number of moles of the solute present per kg of the solvent. If 'n' moles of the solute are dissolved in 'a' kg of the solvent the molality of the solution. Solutions of a particular molality are prepared by dissolving the required number of grams of solute in the required number of kilograms of solvent.

m = $\frac{no\ of\ moles\ of\ solute(n)}{no\ of\ Kg\ of\ the\ solvent(a)}$

but 

n = $\frac{weight\ of\ the\ solute (G)}{gram\ mol\ wt\ of\ the\ solute (GMW)}$.

The fundamental difference between molality and all the other units of concentration is that the amount of solvent is measured, and not the amount of solution. Thus the solutions are not prepared by diluting with the solvent to the specified volume of solution.

Some of the solved problems based on molality are given below:

Question 1: Calculate the molality of a 13% (W/W)solution of H2SO4.
Solution:
13% (W/W) solution means 13g of the solute are present in 100g of solution.

Therefore, weight of the solvent in 100g of solution = (100 - 13) = 87g

Molality of the solution = $\frac{G}{GMW}$ $\times$ $\frac{1000}{'a'\ in\ g}$

= $\frac{13}{98}$ $\times$ $\frac{1000}{87}$ = 1.53m

Question 2: How to prepare a solution in which NaCl must be dissolved in 250.0 grams of water to produce a solution that is 0.20m in NaCl?
Solution:
grams of solute = KgD $\times$ mD $\times$ FWSOL

grams of NaCl = 0.2500 kg $\times$ 0.20m $

grams of NaCl = 2.9 grams

Thus 0.15 grams of NaCl would be weighed and dissolved in 250.0kg of water. The solution would be rendered homogeneous by shaking, swirling or stirring.