Molality Calculator

Molality Calculator

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 = n_solute/m_solvent

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 (m_solu. ) = 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 = n_solute/m_solvent = 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 (m_solu. ): 2025
Solution: 0.21388m

Fig. Concentration Calculator - Molality Calulation Example - Screenshot from CHEMIX School

Dilution

Prepare 200mL of a 2 molal solution by diluting a 5 Molar sodium chloride solution.
FW_NaCl=58.4425g/mol , density of a 2m solution = 1.07417 g/mL , density of water = 1g/mL

Solution:

Convert from molal to molar and solve the problem using:   C₁ V₁ = C₂ V₂

Molarity of a 2molal solution (using 1Kg solvent for simplicity):

m= n/m_solvent     m = 2mol/1Kg     m_solution = 1Kg_solvent+2n = 1000g+ 2*58.4425g = 1116.885g

Volume of solution: V_solution = 1116.885g/1.07417 g/mL = 1039.765mL

M = n/V_solution    C₂ = 2/1.039765Kg = 1.92351M

V₁ = C₂V₂/C₁ = 200mL*1.92354M/5M = 76.94mL

Solution:

Prepare a 2 molal (m) sodium chloride solution by diluting 76.94 mL of the stock concentration (5Molar) solution to 200mL.

CHEMIX School Solution:

1) Select sodium chloride (Substance frame)
2) Insert stock concentration = 5 and select Molar in the Dilution-Stock concentration frame
3) Insert V_solu. = 200mL in the Amount frame
4) Insert 2 in the molality (m) text field (Concentration frame)

Procedure - dilution

      ******** Dilution by volume ********
Put about 90 mL of water into a flask and add 76.9404 mL sodium chloride stock concentration. Add water until 200 mL
       ******** Dilution by mass ********
Put about 90 g of water into a flask and add 91.2324 g of sodium chloride stock concentration. Add water until 214.834 g

Fig. Dilution Calculator - Molality Calulation Example - Screenshot from CHEMIX School
molality calculator dilution example

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