## Molality Calculation

The molality (m)of a solute is the number of moles of that solute divided by the weight of the solvent in kilograms. For water solutions, 1 kg of water has a volume close to that of 1 liter, so molality and molarity are similar in dilute aqueous solutions.
Molality (m) is typically used in thermodynamic calculations where a temperature independent unit of concentration is needed. Since density is a temperature dependent property a solution’s volume, and thus it’s e.g. molar concentration will change as a function of its temperature. By using the solvent’s mass in place of its volume (molal vs molar), the resulting concentration becomes independent of temperature.

Molality (m)  = moles of  solute/kg of solvent = nsolute/msolvent

Solvent :    the liquid in which a solute dissolves
Solute :      the substance that dissolves in a liquid to form a solution
Solution :   is the mixture formed when a solute has dissolved in a solvent

Molality Calculation Example 1

What is the molality (m) of a solution made by mixing 25 g of NaCl and 2.0 Liter of water. Assume the density of water d = 1.0 g/mL.

Solution:
Total mass of solution (msolu. ) = 2kg+0.025kg=2.025Kg

Molar mass of NaCl (FW) = (1 x 22.9898 g/mol ) + (1 x 35.4527 g/mol ) = 58.4425 g

molality(m) = moles of solute/mass of solvent =   nsolute/msolvent = 58.4425/2Kg

m=(25g/mol)/58.4425g= 0,42777mol

0.42777mol/2kg= 0.21388m

CHEMIX School Solution:
1) Select substance: sodium chloride
2) We know the mass of NaCl .  Insert mass : 25
3) Insert the mass of the solution (msolu. )   : 2025
Solution:  0.21388m

Molality Calculation Example 2

Concentrated nitric acid is 70.4% HNO3 by mass (w/w%). What is the  molality of the acid?

Solution:

1) Assuming 100.0 grams of solution:  29.6 g is H2O and 70.4 g is HNO3

2) Calculating the molality:

moles of HNO3 = 70.4 g / 63.0129 g/mol = 1.11723 mol

kg of water = 0.0296 kg

molality = 1.11723 mol / 0.0296 kg = 37.744 m

CHEMIX School Solution
:
1) Select substance: nitric acid
2) In the Concentration frame   (w/w% field) insert:  70.4

Solution:  37.7443m

Molality Calculation Example 3

At 20°C, a 2.32 M aqueous solution of ammonium chloride (NH4Cl) has a density of 1.03439 g/mL What is the molality of ammonium chloride in the solution? The formula weight of NH4Cl is  53.4912g/mol

Solution:

The total mass of  e.g. 1L 2.32M ammonium chloride solution (H2O + NH4Cl) is 1.03439Kg

We have to calculate the masses of both the solute and the solvent:

msolute  (Mass of  NH4Cl)  =  53.4912g/mol * 2.32mol = 124.1g = 0.1241Kg

Mass of  1L 2.32M NH4Cl solution:   1.0344Kgsolution =    msolvent + 0.1241KgNH4Cl

msolvent  =1.0344Kgsolution - 0.1241KgNH4Cl = 0.9103Kgsolvent

Molality Cm =      nsolute/msolvent    =    2.32mol/0.9103Kg = 2.5486m

... or we kan solve this by combining:   CM = nsolute/Vsolution         and         Cm nsolute/msolvent

We can express   nsolute  as      nsolute =  CM*Vsolution

m = (CM*Vsolution)/msolvent  = (2.32mol/L*1L)/(1.0344Kg-2.32mol*((53.4912g/mol)/1000g/kg))

m = 2.32 mol / 0.9103Kg  = 2.5483m

CHEMIX School Solution:
1) Select substance:                              ammonium chloride
2)  In Concentration frame M-field insert :     2.32

Solution:  2.54863m

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