Question
Question: An aqueous solution of \[4\% \] non-volatile solute exerts a pressure of \(1.004bar\) at the normal ...
An aqueous solution of 4% non-volatile solute exerts a pressure of 1.004bar at the normal boiling point of the solvent. What is the molar mass of the solute?
(A)41.35gmol−1
(B)60.21gmol−1
(C)84.42gmol−1
(D)56.91gmol−1
Solution
A solution is made up of solvent and solute. non-volatile solute means which does not evaporate . It has high boiling points and less vapour pressure . we can also say that substances which have boiling point 100∘C or more are termed as volatile rest all are non-volatile .
Complete answer:
Vapour pressure of water at normal boiling point given is 1.013bar
Vapour pressure given for solution in normal boiling point is 1.004bar
Given that solute 4% means 4g
Mass of solvent will be 100−4=96g
Molar mass of water is 18gmol−1
Now we have to apply Raoult’s law
⇒ p10p10−p1=M2×w1w2×M1
Which states that vapour pressure of a solution of non-volatile solute should be equal to the vapour pressure of the pure solvent at that temperature which is multiplied with mole fraction.
Putting all the values
⇒ 1.0131.013−1.004=M2×962×18
M2=84.42gmol−1
Hence the molar mass of the solute is 84.42gmol−1
So the correct answer is Option B.
Additional information:
There are four types of colligative properties which are: vapour pressure, elevation in boiling point, depression in freezing point and osmotic pressure. Colligative is a Latin word which means bounds together.
Colligative properties are those which depend upon the number of solute particles not on the nature of the solution. Basically it does not affect the types of solute present in the solvent only the number of solutes are actually present that matters.
Note:
Remember Raoul’s law only works for lower concentration . Also the interactions of solute and solvent must be identical. As the intermolecular forces vary a lot between the molecules so only by keeping the concentration of solute low we can minimise the intermolecular forces.