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Question: At \({25^ \circ }C\) , the vapor pressure of pure water is \(23.76\,\,mm\,\,of\,\,Hg\) and that of a...

At 25C{25^ \circ }C , the vapor pressure of pure water is 23.76mmofHg23.76\,\,mm\,\,of\,\,Hg and that of a urea solution is 22.98mmofHg22.98\,\,mm\,\,of\,\,Hg . Calculate the molality of the solution.
A) 1.11.1
B) 1.51.5
C) 1.91.9
D) 2.12.1

Explanation

Solution

To solve this question, we must first understand the concepts of Raoult’s Law. Then we need to assess the concepts and formulae to find the molality of the solution and then only we can conclude the correct answer.

Complete solution:
Before we move forward with the solution of this given question, let us first understand some basic concepts regarding Raoult’s Law:
Raoult’s law states that a solvent’s partial vapour pressure in a solution is equal or identical to the vapour pressure of the pure solvent multiplied by its mole fraction in the solution.
Mathematically, Raoult’s law equation is written as;
Psolution=Xsolvent+Psolvent{P_{solution}}\,\, = \,\,{X_{solvent}}\,\, + \,\,{P^ \circ }_{solvent}
Where,
Psolution{P_{solution}} : vapour pressure of the solution
Xsolvent{X_{solvent}}\, : mole fraction of the solvent
Psolvent\,{P^ \circ }_{solvent} : vapour pressure of the pure solvent
Now we will move towards the calculation of the required quantity:
Step 1: According to Raoult’s Law: P=PAXAP = \,\,{P_A}^ \circ {X_A}
23.98=23.76×XA\Rightarrow 23.98 = \,\,23.76 \times {X_A}
XA=0.967\Rightarrow {X_A} = \,\,0.967
Step 2: Mole fraction of Solute: XB=1XA=10.967=0.033{X_B} = \,\,1\,\, - \,\,{X_A} = \,1\,\, - \,\,0.967\,\, = \,\,0.033
Mass of Water =(0.967mol)×(18gmol1)=17.406g = \,\,(0.967\,mol) \times (18\,g\,mo{l^{ - 1}})\,\, = \,\,17.406\,\,g
Now, 17.405g17.405\,\,g of water contains =0.033mole = \,\,0.033\,\,mole of solute
And therefore, 1000g1000\,\,g of water will contain =0.033mol17.406g×1000g== \,\,\dfrac{{0.033\,\,mol}}{{17.406\,\,g}} \times 1000\,\,g\,\, = 1.896mol1.896\,\,mol
Molality is defined as the number of moles of solute present in 1kg1\,\,kg or 1000g1000\,\,g of solvent.
So, clearly we can conclude that, the required molality of the solution is 1.9m1.9\,\,m (approximately)

So, clearly we can conclude that the correct answer is Option C.

Note: The only exception of Raoult’s law is that it applies to solutions. It assumes the ideal behaviour of gases in which the intermolecular forces that are present between dissimilar molecules is zero or non-existent. Meanwhile, Raoult’s law assumes that the intermolecular forces that exist between different molecules and similar molecules are equal. Raoult’s law can also be applied to non-ideal solutions. However, this is done considering the interactions between molecules of different substances.