Example of a Molality Problem: Mole Fraction, Formula and Solution

Example Molality Problem: Mole Fraction, Formula and Solution – On this occasion About the knowledge.co.id will discuss molality with several examples of questions and of course about other things that also cover it. Let's look at the discussion together in the article below to better understand it.

---

Example of a Molality Problem: Mole Fraction, Formula and Solution

---

Molality or molal concentration is a measure of the concentration of a solute in a solution in terms of the amount of substance in a certain mass of the solvent. This is different from the definition of molarity which is based on a certain volume of solution.

The common unit of molality in chemistry is mol/kg. A solution with a concentration of 1 mol/kg is also sometimes expressed as 1 molal. The term molality is formed in analogy to molarity which is the molar concentration of a solution.

The molality of a solution can be tested by adding some solvent. More simply, the Molality (m) of a solution is the moles of solute divided by the kilograms of solvent. Molality or molality is the concentration of a solution which states the number of moles (n) of solute in 1 kg or 1000 grams of solvent.

The difference between molality and molarity is that if molality is molality or molal concentration (m) expresses the number of moles solute in 1000 grams of solvent while molarity is a statement of the number of moles of solute in every one liter solution. Molarity is represented by the notation M and the units are moles/liter.

---

Molality Formula

Information
m = molality (mol/kg)
g = grams of solute (g)
Mm = molar mass of substance (g/mol)
P = mass of solvent (g)

---

Molality Relationship with Mass Percent

Mass percent is a unit of concentration commonly used in chemical solutions. Examples of solutions that we can find everyday are 75% alcohol solution and 24% acetic acid solution. Mass percent is the number of grams of solute in 100 grams of solution mass. The equation that shows the calculation of the mass percent is as follows:

---

Relationship Molality With Molarity

Molarity expresses the number of moles of solute in one liter of solution. Molarity can be converted to molality by converting the volume of the solution into the mass of the solution. Converting volume to mass requires data on the density of the solution (p), which can be formulated as follows:

---

Mole Fraction

The mole fraction is a measure of the concentration of a solution which expresses the ratio of the number of moles in a part of a substance to the total number of moles present in the components of the solution. The mole fraction is divided into 2 parts:

• Solute mole fraction (Xt)

The formula for the mole fraction of the solute (Xt), namely:

Information:
Xt = mole fraction of solute
Nt = number of moles of solute
Np = number of moles of solvent

• Solvent mole fraction (Xp)

The formula for the mole fraction of the solvent (Xp), namely:

Information
Xp = mole fraction of solvent
Nt = number of moles of solute
Np = number of moles of solvent

The sum of the mole fractions of solute and solvent is 1
Xt + Xp = 1

---

Examples of Molality Problems and Solutions

---

Problem 1

What is the molality of a solution containing 4 g of NaOH (Ar Na = 23 g/mol, Ar O = 16 g/mol, and Ar H = 1 g/mol) dissolved in 250 g of water?

Completion:

Is known:

mass of NaOH = 4 gr

ArNa = 23 gr/mol

ArO = 16 g/mol

ArH = 1 gr/mol

mass of water = 250 gr = 0.25 kg

Asked: m = ?

Answer:

Mr. NaOH = 40 gr/mol

number of moles of NaOH = mass/Mr

number of moles of NaOH = 4 gr/(40 gr/mol)

number of moles of NaOH = 0.1 mol

m = number of moles/p

m = 0.1 mol / 0.25 kg

m = 0.4 m

Problem 2

What is the molality of a 37% (w/w) HCl solution? (Ar H = 1 g/mol, Ar Cl = 35.5 g/mol)

Completion:
Is known:

mass of HCl = 37%

ArH = 1 g/mol

ArCl = 35.5 g/mol

Asked: m = ?

Answer:

Mr. HCl = 36.5 gr/mol

suppose the mass of the solution is 100 grams then the mass of HCl is:

HCl mass = 37% x 100 gr

mass of HCl = 37 gr

mass of solvent = mass of solution – mass of HCl

mass of solvent = 100 gr – 37 gr

mass of solvent = 63 gr = 0.063 kg

number of moles of HCl = mass/Mr

number of moles of HCl = 37 gr /(36.5 gr/mol)

number of moles of HCl = 1.01 moles

m = number of moles/mass of solvent

m = 1.01 mol /0.063 kg

m = 16.03 m

Problem 3

Determine the molality of the solution prepared by dissolving 12 grams of urea CO(NH2)2 in 250 grams of water.

Completion:

Is known:

mass of urea = 12 gr

Mr Urea = 60 g/mol

mass of solvent = 250 gr = 0.25 kg

Asked: m = ?

Advertisement

Answer:

number of moles of Urea = mass/Mr

number of moles of Urea = 12 gr /(60 g/mol)

number of moles of Urea = 0.2 moles

m = number of moles/mass of solvent

m = 0.2 mol / 0.25 kg

m = 0.8 m

Problem 4

What is the molality of an alcoholic solution containing 23 mass % ethanol (Mr = 46)?

Answer:

Is known:

mass of ethanol = 23%

Mr ethanol = 46 g/mol

Asked: m = ?

Completion:

for example the mass of the alcohol solution is 100 grams then the mass of ethanol is:

mass of ethanol = 23% x 100 gr

mass of ethanol = 23 gr

mass of solvent = mass of solution – mass of ethanol

mass of solvent = 100 gr – 23 gr

mass of solvent = 77 gr = 0.077 kg

number of moles of ethanol = mass/Mr

number of moles of ethanol = 23 gr /(46 g/mol)

number of moles of ethanol = 0.5 mol

m = number of moles/mass of solvent

m = 0.5 mol /0.077 kg

m = 6.49 m

Problem 5

Calculate the concentration (% by mass) of glucose in a 2 molal glucose solution.

Completion:

Is known:

m = 2 molal = 2 mol/kg = 0.002 mol/gr

Mr glucose = 180 gr/mol

Asked: mass of glucose (%) = ?

Answer:
moles of glucose = mass/Mr

moles of glucose = mass of glucose/(180 g/mol)

moles of glucose = mass of glucose x 0.005 mol/gr

Substitute the moles of glucose into the following equation:

m = moles of glucose/mass of solvent

0.002= mass of glucose x 0.005/mass of solvent

o, oo2/o, oo5 = mass of glucose/mass of solvent

2/5 = mass of glucose/mass of solvent

So the mass ratio of glucose: mass of solvent = 2:5, while the mass ratio of glucose: mass of solution = 2:7.

So,

% mass of glucose = (mass of glucose/mass of solution) x100%

% by mass of glucose = (2/7) x 100%

% glucose mass = 28.57 %

Problem 6

What is the molality of a solution containing 8 g of NaOH (Ar Na = 23 g/mol, Ar O = 16 g/mol, and Ar H = 1 g/mol) dissolved in 250 g of water?

Discussion

Is known:
Mass of NaOH = 8 gr
ArNa = 23 gr/mol
ArO = 16 g/mol
ArH = 1 gr/mol
mass of water = 250 gr = 0.25 kg

Wanted: Molality (m)….?

Answer:
Mr. NaOH = 40 gr/mol
The number of moles of NaOH = mass/Mr
Number of moles of NaOH = 8 gr/(40 gr/mol)
The number of moles of NaOH = 0.2 mol

m = number of moles/p
m = 0.2 mol / 0.25 kg
m = 0.8 m

Problem 6

Determine the molality of the solution prepared by dissolving 15 grams of urea CO(NH2)2 in 250 grams of water.

Discussion

Is known:
Mass of urea = 15 gr
Mr Urea = 60 g/mol
Solvent mass = 250 gr = 0.25 kg

Wanted: Molality (m)…. ?

Answer:
Number of moles of Urea = mass/Mr
Total moles of Urea = 15 gr / 60 g/mol
The number of moles of Urea = 0.25 mol

m = number of moles/mass of solvent
m = 0.25 mol / 0.25 kg
m = 1 m

Problem 7

What is the molality of a 37% (w/w) HCl solution? (Ar H = 1 g/mol, Ar Cl = 35.5 g/mol).

Discussion

Is known:
mass of HCl = 37%
ArH = 1 g/mol
ArCl = 35.5 g/mol

Wanted: Molality (m)…?

Answer:

Mr. HCl = 36.5 gr/mol

Suppose the mass of the solution is 100 grams, then the mass of HCl is:
HCl mass = 37% x 100 gr
mass of HCl = 37 gr

mass of solvent = mass of solution – mass of HCl
mass of solvent = 100 gr – 37 gr
mass of solvent = 63 gr = 0.063 kg

number of moles of HCl = mass/Mr
number of moles of HCl = 37 gr /(36.5 gr/mol)
number of moles of HCl = 1.01 moles

m = number of moles/mass of solvent
m = 1.01 mol /0.063 kg
m = 16.03 m

Problem 8

Determine the amount (grams) of NaOH that must be dissolved in 1 liter of water (water = 1.00 g/mL) to obtain 0.25 m NaOH.

Discussion

Is known:
1 L of water = 1000 mL = 1000 g (since ρ of water = 1.00 g/mL)
mNaOH = 0.25 m
Mr NaOh = 40

Asked: gr…?

Answer:
mNaOH = gr / Mr x 1,000 / P
0.25 = gr/40 x 1,000/1,000
0.25 = g/40
g = 0.25 x 40
g = 10 grams

So, the amount of NaOH needed is 10 grams.

Problem 9

Determine how many mL of water are needed to dissolve 4.9 grams of H2SO4 whose concentration is 0.25 M (Ar H = 1; S = 32; O =16)!
Discussion

Is known:
mH2SO4 = 0.25
Mr H2SO4 = 98
gr = 4.9 grams

Asked :p???

Answer:
m = gr/Mr x 1,000/P
0.25 = 4.9/98 x 1,000/p
p = 20 grams (20 mL)

So, the volume of water is 20 mL.

Problem 10

What mass of water is required to prepare a 1.2 m solution using 0.6 mol NaCl?

Discussion

Problem 11

Suppose there are 2 moles of solute dissolved in 1 liter of solvent, what is the molality?

Discussion

Calculate the molality of 25 grams of KBr (Mr = 119) dissolved in 750 mL of pure water.

Discussion

solute = 25 grams
Mr = 119
solvent (P) = 750 mL = 750 grams (because ρ of water = 1.00 g/mL) = 0.75 kgn =

dissolved substanceMr

n =

25119

= 0.21 mol

m =

nP

m =

0,210,75

= 0.28 m

Thus the review from About the knowledge.co.id about Example of a Molality Problem , hopefully can add to your insight and knowledge. Thank you for visiting and don't forget to read other articles.