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A solution containing $8.44g$ of sucrose in $100g$ of water has a vapour pressure $4.56mm$ of $Hg$ at $273K$. If the vapour pressure of pure water is $4.58mm$ of $Hg$ at the same temperature, calculate the molecular weight of sucrose.

Last updated date: 20th Jun 2024
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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.