Molality Formula

Molality Formula - Molality Definition and Formulas

Molality is defined as the number of moles of solute present in 1000 gm of the solvent. Unlike molarity, molality does not change with temperature since mass is affected by change in temperature. One key thing to note here is that we are using the weight of the ‘Solvent’ and not ‘Solution’ to calculate molality, which can be an easy mistake to commit while solving the problems.
e.g., 1 molal NaCl solution implies 1 mole of NaCl dissolved in 1 kg of water

Mathematically,
${\text{molality}}\left( m \right) = \frac{{{\text{number}}\,{\text{of}}\,{\text{moles}}\,{\text{of}}\,{\text{solute}}\left( n \right)}}{{{\text{weight}}\,{\text{of}}\,{\text{the}}\,{\text{solvent}}\,{\text{in}}\,Kg}}$

Example: Calculate the molality of a solution prepared from 29.22 grams of NaCl in 2.00 kg of water.
Solution:
Solute = 29.22 gm of NaCl
Solvent = 2.00 Kg of water
Molar mass of solute (NaCl) = 58.44 gm/mol
number of moles of solute = $\frac{{29.22}}{{58.44}} = 0.5\,{\text{moles}}$
${\text{molality}}\left( m \right) = \frac{{{\text{number}}\,{\text{of}}\,{\text{moles}}\,{\text{of}}\,{\text{solute}}\left( n \right)}}{{{\text{weight}}\,{\text{of}}\,{\text{the}}\,{\text{solvent}}\,{\text{in}}\,Kg}}$
$m = \frac{{0.5}}{2} = 0.25\,{\text{moles/kg}}$

Question: A 4 g sugar cube (Sucrose: C12H22O11) is dissolved in a 350 ml teacup of 80 °C water. What is the molality of the sugar solution?
Given: Density of water at 80° = 0.975 g/ml
Options:
(a) 0.034 mol/kg
(b) 0.075 mol/kg
(c) 0.089 mol/kg
(d) 0.010 mol/kg