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QJEEAdv19Ch47ec: JEE Advanced 2019 Chemistry - Class 11 JEEAdv19Ch47ec Solution: The mole fraction of urea in an aqueous urea solution…

The mole fraction of urea in an aqueous urea solution containing 900 g of water is 0.05. If the density of the solution is 1.2 g cm −- 3, then molarity of urea solution is ................

(Given data : Molar masses of urea and water are 60 g mol −- 1 and 18 g mol −- 1, respectively)

Step-by-Step Verified Solution

Molarity

(M)=Number of moles of solute×1000Volume of solution (in mL)(M) = {{Number\,of\,moles\,of\,solute \times 1000} \over {Volume\,of\,solution\,(in\,mL)}}

Also, volume = MassDensity{{Mass} \over {Density}}

Given, mole fraction of urea (χurea)({\chi _{urea}}) = 0.05

Mass of water = 900 g

Density = 1.2 g/cm3

χurea{\chi _{urea}} = nureanurea+50{{{n_{urea}}} \over {{n_{urea}} + 50}}

[ ∵\because Moles of water = 90018{{900} \over {18}} = 50]

0.05=0.05 = nureanurea+50{{{n_{urea}}} \over {{n_{urea}} + 50}}

⇒\Rightarrow 19nurea=5019{n_{urea}} = 50

nurea=2.6315{n_{urea}} = 2.6315 moles

wurea=nurea×(M.wt)urea{w_{urea}} = {n_{urea}} \times {(M.wt)_{urea}}

=(2.6315×60)g= (2.6315 \times 60)g

V=2.6315×60+9001.2V = {{2.6315 \times 60 + 900} \over {1.2}}

[ ∵\because Density=Mass of solutionVolume of solutionDensity = {{Mass\,of\,solution} \over {Volume\,of\,solution}} ]

= 881.57 mL

Now, molarity

= Number of moles of solute×1000Volume of solution (mL)Number\,of\,moles\,of\,solute \times {{1000} \over {Volume\,of\,solution\,(mL)}}

=2.6315×1000881.57= {{2.6315 \times 1000} \over {881.57}} = 2.98 M

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