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1800-102-2727Rohit, who is in class 11, signed up for swimming lessons. He had no prior swimming experience, so at first, he could not swim well, but over time he grew to enjoy swimming and could swim well.
After a few months of consistent practice, he suddenly became aware that his skin was looking darker than usual.
He consulted his older sister who is a dermatologist to learn the reason. She questioned Rohit about whether he had ever checked the pool's chlorine level, adding that tanning is caused by exposure to too much chlorine.
But how can we determine the chlorine level in the pool? How do we know if the pool water has too much chlorine?
The amount of chlorine in water can be expressed using the concentration terms like molality, molarity, ppm, etc. If the concentration of chlorine in the pool water is greater than 1-3 ppm (parts per million), it is said to be in excess. A solution is made up of solute and solvent. In this scenario, chlorine is the solute and water is the solvent. The amount of chlorine in water is basically the level or concentration of chlorine in the water.
Let’s get to know more about the concept of molality and its relationship with other concentration terms.
TABLE OF CONTENTS
Molality is defined as the “total number of moles of a solute present in a kilogram of a solvent.”
Molal concentration is another name for molality. It is a measurement of a solution's solute concentration. The two components that make up the solution are the solute and the solvent. The terms parts per million, weight percentage, volume percentage, formality, normality, molality and molarity are just a few of the various methods to indicate the concentration of a solution.
Mathematically, molality can be expressed as
$Molality\left(m\right)=\frac{Numberofmolesofthesolute\left(mol\right)}{Massofthesolvent\left(kg\right)}$
Units
$Molality\left(m\right)=\frac{moles}{kg}=moles/kg$
For instance, if one mole of sugar is dissolved in one kilogram of water, the resulting solution would not be 1 molal or 1 m.
Mass of the Solvent
Molality is inversely proportional to the mass of the given solvent. If the mass of the solvent is increased keeping the number of moles of the solute constant, molality decreases.
$Molality\propto \frac{1}{massofthesolvent}$
Moles of the Solute
Molality is directly proportional to the number of moles of the solute. If the number of moles of the solute is increased keeping the mass of the solvent constant, molality increases.
$Molality\propto Numberofmolesofthesolute$
Temperature
Unlike molarity, molality depends on the mass of the solvent. Mass is a physical quantity which is independent of temperature. Hence, molality is temperature-independent.
Both molarity and molality are very useful concentration expressions. They could also be connected to the solution's density in order to change from one to the other. The derivation of the relationship is given below.
Consider the mass of the given solute to be ${W}_{1}g=\frac{{W}_{1}}{10000}kg$
The solute has a molar mass of M1 g.
The solvent is assumed to weigh ${W}_{2}g=\frac{{W}_{2}}{1000}kg$
The volume of the prepared solution is assumed to be $VmL=\frac{V}{1000}L$
We know that,
$Numberofmoles\left(X\right)ofsolute=\frac{massofthesolute}{molarmassofthesolute}=\frac{{W}_{1}}{{M}_{1}}...............\left(1\right)$
According to the above formula, molarity can be expressed as
$Molarity\left(M\right)=\frac{NumberofMolesofthesolute}{Volumeofthesolution(inlitres)}=\frac{{W}_{1}}{{M}_{1}}/\frac{V}{1000}=\frac{{W}_{1}\times 1000}{{M}_{1}\times V}............\left(2\right)$
Similarly,
$Molality\left(m\right)=\frac{NumberofMolesofthesolute}{Massofthesolvent(inkg)}=\frac{{W}_{1}}{{M}_{1}}/\frac{{W}_{2}}{1000}=\frac{{W}_{1}\times 1000}{{M}_{1}\times {W}_{2}}...............\left(3\right)$
From equation (2),
$V=\frac{{W}_{1}\times 1000}{{M}_{1}\times M}$
From equation (3),
${W}_{2}=\frac{{W}_{1}\times 1000}{{M}_{1}\times m}$
$Density,\rho =\frac{massofsoluteandsolvent}{Volumeofsolvent}$
$\rho =\frac{{W}_{1}+{W}_{2}}{V}...............\left(4\right)$
${W}_{1}+{W}_{2}={W}_{1}+\frac{{W}_{1}\times 1000}{{M}_{1}\times m}$
Multiplying and dividing by 1000M1 on both numerator and denominator of W1,
$\frac{{W}_{1}\times 1000\times {M}_{1}}{1000\times {M}_{1}}+\frac{{W}_{1}\times 1000}{{M}_{1}\times m}$
${W}_{1}+{W}_{2}=\frac{{W}_{1}\times 1000}{{M}_{1}}(\frac{{M}_{1}}{1000}+\frac{1}{m})$
Substituting the value of W1+W2 and of V in equation (4),
$\rho =\frac{\frac{{W}_{1}\times 1000}{{M}_{1}}(\frac{{M}_{1}}{1000}+\frac{1}{m})}{\frac{W\times 1000}{{M}_{1}\times M}}$
$\rho =\frac{(\frac{{M}_{1}}{1000}+\frac{1}{m})}{\frac{1}{M}}$
$\rho =(\frac{{M}_{1}}{1000}+\frac{1}{m})\times M$
Density of solution $\left(\rho \right)=\frac{M{M}_{1}}{1000}+\frac{M}{m}$
$\frac{M}{m}=\rho -\frac{M{M}_{1}}{1000}$
Molarity and molality are related as shown in the equation above.
$Molality\left(m\right)=\frac{NumberofMolesofthesolute}{Massofthesolvent(inKg)}$
$Molality\left(m\right)=\frac{Massofthesolute}{Molarmassofthesolute}\times \frac{1}{Massofthesolvent(inKg)}$
$Masspercent=(\frac{Massofthesolute}{Massofthesolution}\times 100)\%$
We know that,
$Numberofmoles=\frac{massofsolute}{molarmassofsolute}$
$Massofsolute=\frac{masspercent}{100}\times massofsolution$
Substituting the above formula in the formula of molality,
$Molality\left(m\right)=\frac{\frac{masspercent}{100}\times massofsolution}{molarmassofsolute}\times \frac{1}{Massofthesolvent(inKg)}$
$Molality=\frac{masspercent\times massofsolution}{Massofthesolvent(inKg)\times 100\times molarmassofsolute}$
Let’s assume a solution where the number of moles of the solute is nB and the number of moles of the solvent is nA. The weight of the solvent is assumed to be wA g and the molar mass of the solvent is MA g.
Mole fraction is defined as the ratio of the number of moles of a certain component in a mixture and the total number of moles in that given mixture.
Mole fraction is expressed mathematically as:
${\chi}_{A}=\frac{{n}_{A}}{{n}_{A}{+n}_{B}}$ …………….. (5)
${\chi}_{B}=\frac{{n}_{B}}{{n}_{A}{+n}_{B}}$ ………………(6)
$\frac{{\chi}_{A}}{{\chi}_{B}}=\frac{{n}_{B}}{{n}_{A}}$ ……………….(7)
$Molality\left(m\right)=\frac{Numberofmolesofthesolute}{Massofthesolvent(inkg)}$
As the mass of the solute is present in g, the above formula becomes
$m=\frac{{n}_{B}}{{w}_{A}}\times 1000$
${n}_{B}=\frac{m\times {w}_{A}}{1000}$
Substituting the value of nB in equation (7),
$\frac{{\chi}_{B}}{{\chi}_{A}}=\frac{\frac{m\times {w}_{A}}{1000}}{{n}_{A}}=\frac{m\times {w}_{A}}{1000}\times \frac{1}{{n}_{A}}$………………….(8)
We know that,
$Numberofmoles=\frac{massofsolute}{molarmassofsolute}$
${n}_{A}=\frac{{w}_{A}}{{M}_{A}}$
Substituting the value of nA in equation (8),
$\frac{{\chi}_{B}}{{\chi}_{A}}=\frac{m\times {w}_{A}}{1000}\times \frac{1}{{w}_{A}}\times {M}_{A}$
$\frac{{\chi}_{B}}{{\chi}_{A}}=\frac{m}{1000}\times {M}_{A}$
$m=\frac{{\chi}_{B}}{{\chi}_{A}}\times \frac{1000}{{M}_{A}}$ ………………..(9)
We know that, ${\chi}_{A}+{\chi}_{B}=1$
${\chi}_{B}=1-{\chi}_{A}$
Substituting the value of B in equation (9),
m= B1-B 1000MA
The above equation states the relationship between molality and mole fraction.
Parameters |
Molarity |
Molality |
Definition |
Molarity is the number of moles of a solute in a litre of a solution. Molarity is also referred to as a solution's molar concentration. |
Molal concentration is another name for molality. Molality is defined as the “total number of moles of a solute present in a kilogram of a solvent.” |
Representation |
M |
m |
Formula |
$Molarity\left(M\right)=\frac{NumberofMolesofthesolute}{Volumeofthesolution(inlitres)}$ |
$Molality\left(M\right)=\frac{NumberofMolesofthesolute}{Massofthesolvent(inkg)}$ |
Temperature dependency |
Molarity is temperature-dependent. It varies inversely with temperature. Molarity (M) 1Volume 1Temperature |
Molality is temperature-independent as it depends on the mass of the solvent and not on the volume of the solvent. |
Unit |
mol L^{-1} |
mol kg^{-1} |
1. Calculate the molality of a 2 M CH3COOH solution if its density is 1.21 gL^{-1}.
a. 2.50 m
b. 1.50 m
c. 1.30 m
d. 1.40 m
Answer: B
Solution: We know, density of solution $\left(\rho \right)=\frac{M{M}_{1}}{1000}+\frac{M}{m}$
Molar mass of CH_{3}COOH = 60 g mol^{-1}
$\rho =\frac{M{M}_{1}}{1000}+\frac{M}{m}=\frac{2\times 60}{1000}+\frac{2}{m}$
$\rho =0.12+\frac{2}{m}$
$\frac{2}{m}=1.21-0.12$
$m=\frac{2}{1.33}\approx 1.50$
So, option B is the correct answer.
2. Find the molality of an aqueous solution of glucose having a mole fraction of 0.2 and a density of 1.1 g mol^{-1}.
a. 13.88 m
b. 18.33 m
c. 6.24 m
d. 2.64 m
Answer: A
Molar mass of the solute (glucose) (MB) = 180 g mL^{-1}
Molar mass of the solvent (water) (MA) = 18 g mL^{-1}
Mole fraction of glucose (B)=0.2
Mole fraction of water (A)=0.8
Mass of the solute (glucose) = WB
Mass of the solvent (water) = WA
Mole fraction of glucose ${(\chi}_{B})=\frac{\frac{{W}_{B}}{{M}_{B}}}{\frac{{W}_{A}}{{M}_{A}}+\frac{{W}_{B}}{{M}_{B}}}$
${(\chi}_{B})=\frac{\frac{{W}_{B}}{180}}{\frac{{W}_{A}}{18}+\frac{{W}_{B}}{180}}$
$0.2=\frac{\frac{{W}_{B}}{180}}{10{W}_{A}+{W}_{B}}$
${W}_{B}=\frac{5}{2}{W}_{A}$
$Molality\left(m\right)=\frac{Numberofmolesofthesolute\left(mol\right)}{Massofthesolvent\left(kg\right)}$
$Numberofmoles=\frac{massofsolute}{molarmassofsolute}=\frac{{W}_{B}}{{M}_{B}}$
$Molality\left(m\right)=\frac{{W}_{B}}{{M}_{B}}\times \frac{1000}{MassoftheSolvent\left(kg\right)}$
$Molality\left(m\right)=\frac{{5W}_{A}}{2{M}_{B}}\times \frac{1000}{{W}_{A}}$
$Molality\left(m\right)=13.88m$
So, option A is the correct answer.
3. Which of the following will alter as the temperature increases?
a. Molality
b. Molarity
c. Both of these
d. None of these
Answer: B
Solution: Molarity is a temperature-dependent concentration term as it varies inversely with temperature. Mathematically, $Molarity\left(M\right)\alpha \frac{1}{Volume}\alpha \frac{1}{Temperature}$
So, option B is the correct answer.
4. What is the unit of molarity?
a. mol L^{-1}
b. mol Kg^{-1}
c. mol^{-1}L^{-1}
d. mol L
Answer: A
Solution: The number of moles of a solute in a specific volume of a solution is known as its molarity (M). In other words, molarity is the number of moles of a solute in a litre of a solution. Molarity is also referred to as a solution's molar concentration.
Mathematically,
$Molarity\left(M\right)=\frac{NumberofMolesofthesolute}{Volumeofthesolution(inlitres)}$
Hence, the unit of molarity = molL^{-1}
So, option A is the correct answer.
1. Is molality an extensive or intensive property?
Answer: A property of matter that does not change as the amount of the substance changes is called an intensive property. Given that it is a bulk property, it is a physical characteristic independent of the size or mass of a sample. An extensive property, on the other hand, is one that is sample size dependent. It includes properties like mass and volume. The ratio of two extensive properties is an intensive property. For instance, mass per unit volume is density (an intensive property), whereas mass and volume are both extensive properties. As both moles and mass are extensive properties, molality is the ratio of two extensive properties. Hence, molality is an intensive property.
2. Can the value of molarity and molality of an infinitely diluted solution be equated to 1?
Answer: For the sake of numerical simplicity, we can only consider molarity = molality for an extremely diluted aqueous solution. Water has a known density of 1 g cc^{-1}. As a result, in extremely diluted solutions, the solute's weight becomes almost negligible. The 1 L volume of the solution now equals 1 kg. You can think of the solvent's weight as being equal to the weight of the solution.
3. How can the concentration of a solution be altered?
Answer: The concentration of a solution can be altered by adjusting the amount of solvent. The concentration of a solute in a solution decreases when more solvent is introduced. Generally, on adding more solvent, the concentration of the solution decreases, and on removing solvent, the concentration of the solution increases.
4. How do you make a solution with a known concentration?
Answer: By either dissolving a known mass of the solute and diluting it to the desired final volume, or by diluting the acceptable volume of the more concentrated solution to the needed final volume, solutions with known concentrations can be prepared.