Molality Calculation
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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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