## 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 = **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*

**Molality Calculation Example 2**

Concentrated nitric acid is 70.4% HNO_{3} by mass
(w/w%). What is the molality of the acid?

**Solution:**

1) Assuming
100.0 grams of solution: 29.6 g is H_{2}O and 70.4 g
is HNO_{3}

2) Calculating
the molality:

moles of HNO_{3} =
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 (NH_{4}Cl)
has a density of 1.03439 g/mL What is the molality of ammonium
chloride in the solution? The formula weight of NH_{4}Cl
is 53.4912g/mol

**Solution****:**

The total mass of e.g. 1L 2.32M ammonium chloride solution (H_{2}O
+ NH_{4}Cl) is 1.03439Kg

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

m_{solute} (Mass of NH_{4}Cl)
= 53.4912g/mol * 2.32mol = 124.1g = 0.1241Kg

Mass of 1L 2.32M NH_{4}Cl solution: 1.0344Kg_{solution}
= m_{solvent} + 0.1241Kg_{NH}_{4}_{Cl}_{
}

m_{solvent} =1.0344Kg_{solution} - 0.1241Kg_{NH4Cl}
= 0.9103Kg_{solvent}

Molality **C**_{m} = **n**_{solute}/**m**_{solvent}
= 2.32mol/0.9103Kg = **2.5486m**

... or we kan solve this by combining: C_{M} =
n_{solute}/V_{solution}
and ** C**_{m}
= _{ }**n**_{solute}/**m**_{solvent}

We can express ** n**_{solute} as
** n**_{solute} = C_{M}***V**_{solution}

**m** = (**C**_{M}***V**_{solution})/**m**_{solvent}
= (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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