Answer
437.4k+ views
Hint: By examining our question write the set of data given to find the number of moles then the mole fraction of sucrose then gets the molecular mass of sucrose by relative lowering of vapour pressure which is equal to the mole fraction of solute.
Complete answer:
We have learnt that vapour pressure of solution decreases when a non-volatile solute is added to a volatile solvent. There are many properties of solutions which are connected with this property. There are relative lowering of vapour pressure of the solvent, depression of freezing point of the solvent, elevation of boiling point of the solvent and osmotic pressure of solution these all properties which we have named till now depend on the number of solute particles and they are not depended on their nature but they do depend on the total number of particles present in the solution and we know such properties are called as colligative properties.
And now when we talk about Relative lowering of vapour pressure which is expressed in an equation form as $\begin{align}
& \dfrac{\Delta p}{p}=\frac{4.58-4.56}{4.58}=\dfrac{1.52}{M} \\
& M=3.48 g/mol \\
& =\dfrac{100}{8}=5.56 \\
& =18g/mol \\
& =\dfrac{\dfrac{8.44}{M}}{\frac{8.44}{M}+5.56}\simeq \dfrac{8.44}{M\times 5.56}\simeq \dfrac{1.52}{M} \\
\end{align}$ .
Now let’s go back to our question from which lets first determine some values
Let molar weight of sucrose $=M$$g/mol$
Then the number of moles of sucrose will be $=\dfrac{8.44}{M}$
Now the molecular weight of water will be $=18g/mol$
Then the number of moles of water will be $=\dfrac{100}{8}=5.56$
Now let us find the mole fraction of sucrose $=\dfrac{\dfrac{8.44}{M}}{\frac{8.44}{M}+5.56}\simeq \dfrac{8.44}{M\times 5.56}\simeq \dfrac{1.52}{M}$
So now as we know the relative lowering of vapour pressure is equal to the mole fraction of solute so we can write the same in the form of equation as
$\dfrac{\Delta p}{p}=\dfrac{p{}^\circ -p}{p{}^\circ }={{X}_{solute}}$
Now putting the values, we will get
$\dfrac{\Delta p}{p}=\dfrac{4.58-4.56}{4.58}=\dfrac{1.52}{M}$
Now by solving this equation we get
$M=3.48$ $g/mol$
Hence the molecular weight of sucrose is $3.48g/mol$.
Note: The point which we should know is that the lowering of vapour pressure depends only on the concentration of the solute particles and is not dependent on their identity which was established by Raoult. One more point which we need to consider is that in a solution which contains several non-volatile solutes in that the lowering of the vapour pressure depends on the sum of the mole fraction of different solute which we have used in our solution above.
Complete answer:
We have learnt that vapour pressure of solution decreases when a non-volatile solute is added to a volatile solvent. There are many properties of solutions which are connected with this property. There are relative lowering of vapour pressure of the solvent, depression of freezing point of the solvent, elevation of boiling point of the solvent and osmotic pressure of solution these all properties which we have named till now depend on the number of solute particles and they are not depended on their nature but they do depend on the total number of particles present in the solution and we know such properties are called as colligative properties.
And now when we talk about Relative lowering of vapour pressure which is expressed in an equation form as $\begin{align}
& \dfrac{\Delta p}{p}=\frac{4.58-4.56}{4.58}=\dfrac{1.52}{M} \\
& M=3.48 g/mol \\
& =\dfrac{100}{8}=5.56 \\
& =18g/mol \\
& =\dfrac{\dfrac{8.44}{M}}{\frac{8.44}{M}+5.56}\simeq \dfrac{8.44}{M\times 5.56}\simeq \dfrac{1.52}{M} \\
\end{align}$ .
Now let’s go back to our question from which lets first determine some values
Let molar weight of sucrose $=M$$g/mol$
Then the number of moles of sucrose will be $=\dfrac{8.44}{M}$
Now the molecular weight of water will be $=18g/mol$
Then the number of moles of water will be $=\dfrac{100}{8}=5.56$
Now let us find the mole fraction of sucrose $=\dfrac{\dfrac{8.44}{M}}{\frac{8.44}{M}+5.56}\simeq \dfrac{8.44}{M\times 5.56}\simeq \dfrac{1.52}{M}$
So now as we know the relative lowering of vapour pressure is equal to the mole fraction of solute so we can write the same in the form of equation as
$\dfrac{\Delta p}{p}=\dfrac{p{}^\circ -p}{p{}^\circ }={{X}_{solute}}$
Now putting the values, we will get
$\dfrac{\Delta p}{p}=\dfrac{4.58-4.56}{4.58}=\dfrac{1.52}{M}$
Now by solving this equation we get
$M=3.48$ $g/mol$
Hence the molecular weight of sucrose is $3.48g/mol$.
Note: The point which we should know is that the lowering of vapour pressure depends only on the concentration of the solute particles and is not dependent on their identity which was established by Raoult. One more point which we need to consider is that in a solution which contains several non-volatile solutes in that the lowering of the vapour pressure depends on the sum of the mole fraction of different solute which we have used in our solution above.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science 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)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 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)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![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)
How do you graph the function fx 4x class 9 maths 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)