
Derive the relation between molarity and molality. Calculate the molality of 1M HCl solution having density ${ 1.5365{ gmol }^{ -1 } }$.
Answer
585k+ views
Hint: Molarity is also known as molar concentration. It is the concentration of the solution. It is the number of moles of solute per liter of solution.
Molality: It is defined as the number of moles of solute in one kilogram of solvent.
Complete step by step solution:
The formulas of molarity and molality are given below:
${ Molarity= }\dfrac { moles\quad of\quad solute }{ liters\quad of\quad solution } $
and ${ Molality= }\dfrac { moles\quad of\quad solute }{ kilograms\quad of\quad solvent } $
Now, the relationship between Molarity and Molarity:
Let,
The mass of given solute = W
The volume of the solution = V
Molality = m
The molar mass of solute = M
Molarity = M
Therefore, molarity, M will be = $\dfrac { W }{ M^{ ' } } { \times \dfrac { 1000 }{ V } }$.........(1)
And molality, m will be = $\dfrac { W }{ M^{ ' } } { \times \dfrac { 1000 }{ { W }^{ ' } } }$………..(2)
As we know, Density = mass/volume
${ d= }\dfrac { M }{ V } $
From here, ${ d= }\dfrac { W+W^{ ' } }{ V } $
Now, from equation (1), we get
${ V= }\dfrac { W\times 1000 }{ m\times M^{ ' } } $ .........(3)
And from equation (2), we get
${ { W }^{ ' }= }\dfrac { W\times 1000 }{ m\times M^{ ' } } $
Therefore,
${ W+{ W }^{ ' }= }{ W+ }\dfrac { W\times 1000 }{ m\times M^{ ' } } { \times \dfrac { 1000 }{ { M }^{ ' } } }\left[ \dfrac { { mM }^{ ' } }{ 1000m } +\dfrac { 1 }{ m } \right] { W }^{ ' }$.........(4)
Now, by dividing equation (4) by (3), we get
$\dfrac { d }{ m } { = }\dfrac { 1 }{ m } { + }\dfrac { { M }^{ ' } }{ 1000 } \dfrac { 1 }{ m } $
$\dfrac { d }{ m } { - }\dfrac { { M }^{ ' } }{ 1000 } { \times m }$
$\dfrac { M\times 1000 }{ \left( d\times 1000 \right) -M\left( { M }^{ ' } \right) } $
Hence, this is the relation between molality and molarity.
NUMERICAL:
It is given that,
Density = ${ 1.5365{ gmol }^{ -1 } }$.
Molecular weight of solute (HCl) = ${ 36.46{ gmol }^{ -1 } }$.
Molarity = 1M
Molality, m =?
Using the formula;
${ Molality=\dfrac { molarity }{ \left( density-molarity \right) \times molecular\quad weight\quad of\quad solute } }$
Now, by putting the values in the above formula, we get
Molality = ${ Molality=\dfrac { 1 }{ { \left( { 1.5365gmol }^{ -1 }{ -1 } \right) \times 36.46 } } }$
Molality = ${ Molality=\dfrac { 1 }{ { 0.5365\times 36.46 } } }$
= ${ Molality=\dfrac { 1 }{ { 19.560 } } =0.05112 }$
Hence, the molality of 1M HCl solution = ${ 0.05112M }$.
Note: The possibility to make a mistake is that don’t make mistakes in the values given in the question or any error using formulas or analysis. Don’t confuse between molality and molarity also convert the values from kg to g in molality.
Molality: It is defined as the number of moles of solute in one kilogram of solvent.
Complete step by step solution:
The formulas of molarity and molality are given below:
${ Molarity= }\dfrac { moles\quad of\quad solute }{ liters\quad of\quad solution } $
and ${ Molality= }\dfrac { moles\quad of\quad solute }{ kilograms\quad of\quad solvent } $
Now, the relationship between Molarity and Molarity:
Let,
The mass of given solute = W
The volume of the solution = V
Molality = m
The molar mass of solute = M
Molarity = M
Therefore, molarity, M will be = $\dfrac { W }{ M^{ ' } } { \times \dfrac { 1000 }{ V } }$.........(1)
And molality, m will be = $\dfrac { W }{ M^{ ' } } { \times \dfrac { 1000 }{ { W }^{ ' } } }$………..(2)
As we know, Density = mass/volume
${ d= }\dfrac { M }{ V } $
From here, ${ d= }\dfrac { W+W^{ ' } }{ V } $
Now, from equation (1), we get
${ V= }\dfrac { W\times 1000 }{ m\times M^{ ' } } $ .........(3)
And from equation (2), we get
${ { W }^{ ' }= }\dfrac { W\times 1000 }{ m\times M^{ ' } } $
Therefore,
${ W+{ W }^{ ' }= }{ W+ }\dfrac { W\times 1000 }{ m\times M^{ ' } } { \times \dfrac { 1000 }{ { M }^{ ' } } }\left[ \dfrac { { mM }^{ ' } }{ 1000m } +\dfrac { 1 }{ m } \right] { W }^{ ' }$.........(4)
Now, by dividing equation (4) by (3), we get
$\dfrac { d }{ m } { = }\dfrac { 1 }{ m } { + }\dfrac { { M }^{ ' } }{ 1000 } \dfrac { 1 }{ m } $
$\dfrac { d }{ m } { - }\dfrac { { M }^{ ' } }{ 1000 } { \times m }$
$\dfrac { M\times 1000 }{ \left( d\times 1000 \right) -M\left( { M }^{ ' } \right) } $
Hence, this is the relation between molality and molarity.
NUMERICAL:
It is given that,
Density = ${ 1.5365{ gmol }^{ -1 } }$.
Molecular weight of solute (HCl) = ${ 36.46{ gmol }^{ -1 } }$.
Molarity = 1M
Molality, m =?
Using the formula;
${ Molality=\dfrac { molarity }{ \left( density-molarity \right) \times molecular\quad weight\quad of\quad solute } }$
Now, by putting the values in the above formula, we get
Molality = ${ Molality=\dfrac { 1 }{ { \left( { 1.5365gmol }^{ -1 }{ -1 } \right) \times 36.46 } } }$
Molality = ${ Molality=\dfrac { 1 }{ { 0.5365\times 36.46 } } }$
= ${ Molality=\dfrac { 1 }{ { 19.560 } } =0.05112 }$
Hence, the molality of 1M HCl solution = ${ 0.05112M }$.
Note: The possibility to make a mistake is that don’t make mistakes in the values given in the question or any error using formulas or analysis. Don’t confuse between molality and molarity also convert the values from kg to g in molality.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

What is periodicity class 11 chemistry CBSE

Explain zero factorial class 11 maths CBSE

What is a periderm How does periderm formation take class 11 biology CBSE

What are porins class 11 biology CBSE

