Answer
384k+ views
Hint: By using the formula of depression in freezing point one can determine the relation between the freezing point and the molecular weight. The depression in freezing point is equal to the product of cryoscopic constant and molality.
Complete step by step answer:
The freezing point depression is explained as the lowering of the freezing point of the solvent with the addition of the solute.
The freezing point depression is the colligative property which is proportional to the molality of the solute added.
The freezing point depression is given by the formula as shown below.
$\Delta {T_f} = {K_f} \times m$
Where,
$\Delta {T_f}$ is the freezing point depression
${K_f}$ is the cryoscopic constant
m is the molality
The molality is defined as the number of moles of solutes dissolved in one kilogram of solvent.
The formula of molality is shown below.
$m = \dfrac{n}{{{M_1}}}$
Where,
m is the molality
n is the number of moles
${M_1}$ is the mass in kilogram
The number of moles is given by the formula as shown below.
$n = \dfrac{m}{{{M_2}}}$
Where,
n is the number of moles
m is the mass
${M_2}$ is the molecular weight
So, if we substitute the terms in the formula of freezing point depression then the new formula obtained will be.
$ \Rightarrow \Delta {T_f} = {K_f} \times \dfrac{m}{{{M_2}{M_1}}}$
If we keep the ${K_f}$, m and ${M_1}$ as constant then the freezing point depression will be inversely proportional to the molecular weight.
$ \Rightarrow \Delta {T_f} = \dfrac{1}{{{M_2}}}$
So, by increasing the value of depression in freezing point, the molecular weight decreases and vice versa.
Thus, an increase in the molecular weight will have a smaller effect on the freezing point.
Note:
The freezing point is calculated by subtracting the depression in freezing point value and the freezing point of pure water which is zero degree Celsius.
Complete step by step answer:
The freezing point depression is explained as the lowering of the freezing point of the solvent with the addition of the solute.
The freezing point depression is the colligative property which is proportional to the molality of the solute added.
The freezing point depression is given by the formula as shown below.
$\Delta {T_f} = {K_f} \times m$
Where,
$\Delta {T_f}$ is the freezing point depression
${K_f}$ is the cryoscopic constant
m is the molality
The molality is defined as the number of moles of solutes dissolved in one kilogram of solvent.
The formula of molality is shown below.
$m = \dfrac{n}{{{M_1}}}$
Where,
m is the molality
n is the number of moles
${M_1}$ is the mass in kilogram
The number of moles is given by the formula as shown below.
$n = \dfrac{m}{{{M_2}}}$
Where,
n is the number of moles
m is the mass
${M_2}$ is the molecular weight
So, if we substitute the terms in the formula of freezing point depression then the new formula obtained will be.
$ \Rightarrow \Delta {T_f} = {K_f} \times \dfrac{m}{{{M_2}{M_1}}}$
If we keep the ${K_f}$, m and ${M_1}$ as constant then the freezing point depression will be inversely proportional to the molecular weight.
$ \Rightarrow \Delta {T_f} = \dfrac{1}{{{M_2}}}$
So, by increasing the value of depression in freezing point, the molecular weight decreases and vice versa.
Thus, an increase in the molecular weight will have a smaller effect on the freezing point.
Note:
The freezing point is calculated by subtracting the depression in freezing point value and the freezing point of pure water which is zero degree Celsius.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)