
What is Raoult’s law? How can molar mass of a non-volatile solute be determined with its help?
Answer
598.8k+ views
Hint: Raoult’s law is only valid for ideal solutions. A non- volatile solute is a substance that does not readily evaporate into a gas under existing conditions. And a non-volatile solute exhibits a low vapor pressure and a high boiling point. For example sugar.
Complete step-by-step answer:
Raoult’s law states that the relative lowering of vapor pressure of a solution containing a non-volatile solute is equal to the mole fraction of the solute in the solution, it means that vapor pressure of a solution is dependent on the mole fraction of a solute added to the solution. And from the definition we can derive the formula of Raoult’s law:
\[{{P}_{solution}}={{X}_{solvent}}P_{solvent}^{0}\]
Where, \[{{P}_{solution}}\] is the vapor pressure of solution,
\[{{X}_{solvent}}\] is the mole fraction of solvent = moles of solvent / total number of moles
\[=\dfrac{{{n}_{solvent}}}{{{n}_{totalmoles}}}\]
\[P_{solvent}^{0}\]is the vapor pressure of pure solvent at a particular temperature.
The molar mass of a non-volatile solute can be determined by using following equation:
And here we know that
\[\Delta P={{P}_{0}}-P\],
and \[{{X}_{solute}}={{X}_{2}}=\dfrac{\tfrac{{{W}_{2}}}{{{M}_{2}}}}{\tfrac{{{W}_{1}}}{{{M}_{1}}}+\tfrac{{{W}_{2}}}{{{M}_{2}}}}\]
By putting the value of \[{{X}_{solute}}\] and \[\Delta P\] and then rearranging the equation we get:
And here;
\[{{P}_{0}}\] = vapor pressure of pure solvent,
\[P\] = vapor pressure of solution,
\[{{M}_{2}}\] = mass of solute,
\[{{W}_{2}}\] =molar mass of solute,
\[{{M}_{1}}\] = mass of solvent,
\[{{W}_{1}}\] = molar mass of solvent.
So, from this derivation we can find the molar mass of non-volatile solute (\[{{W}_{2}}\]), if the other data is given.
Note: We should be care full about the units. And Raoult’s law is applicable for very dilute solutions and for solutions containing non-volatile solute only. Raoult’s law is not applicable to solutes which dissociate or associate in the particular solution.
Complete step-by-step answer:
Raoult’s law states that the relative lowering of vapor pressure of a solution containing a non-volatile solute is equal to the mole fraction of the solute in the solution, it means that vapor pressure of a solution is dependent on the mole fraction of a solute added to the solution. And from the definition we can derive the formula of Raoult’s law:
\[{{P}_{solution}}={{X}_{solvent}}P_{solvent}^{0}\]
Where, \[{{P}_{solution}}\] is the vapor pressure of solution,
\[{{X}_{solvent}}\] is the mole fraction of solvent = moles of solvent / total number of moles
\[=\dfrac{{{n}_{solvent}}}{{{n}_{totalmoles}}}\]
\[P_{solvent}^{0}\]is the vapor pressure of pure solvent at a particular temperature.
The molar mass of a non-volatile solute can be determined by using following equation:
And here we know that
\[\Delta P={{P}_{0}}-P\],
and \[{{X}_{solute}}={{X}_{2}}=\dfrac{\tfrac{{{W}_{2}}}{{{M}_{2}}}}{\tfrac{{{W}_{1}}}{{{M}_{1}}}+\tfrac{{{W}_{2}}}{{{M}_{2}}}}\]
By putting the value of \[{{X}_{solute}}\] and \[\Delta P\] and then rearranging the equation we get:
And here;
\[{{P}_{0}}\] = vapor pressure of pure solvent,
\[P\] = vapor pressure of solution,
\[{{M}_{2}}\] = mass of solute,
\[{{W}_{2}}\] =molar mass of solute,
\[{{M}_{1}}\] = mass of solvent,
\[{{W}_{1}}\] = molar mass of solvent.
So, from this derivation we can find the molar mass of non-volatile solute (\[{{W}_{2}}\]), if the other data is given.
Note: We should be care full about the units. And Raoult’s law is applicable for very dilute solutions and for solutions containing non-volatile solute only. Raoult’s law is not applicable to solutes which dissociate or associate in the particular solution.
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