Question

An aqueous solution of urea containing 18g urea in 15cm³ of solution having density equal to 1.052g/cm³.If the molecular weight of urea is 60.Then molality of the solution is

Answer 1

Hello Dear.

There is the little typing error in the Question. The Volume of the Aqueous Solution of Urea must be 1500 cm³ or the mass of the urea must be 15 g or less than that.

Now, as per as the Volume 1500 cm³, solution is shown.

 Volume of the Aqueous Solution = 1500 cm³
Density of the Solution = 1.052 g/cm³

∴ Mass of the solution = Density × Volume.
    = 1.052 × 1500
    = 1578 g.

Mass of the solvent = Mass of the Solution - Mass of the solute (or Urea).
= 1578 - 18
= 1560 g.
= 1.56 kg.

No. of moles = Mass/Molar Mass
∴No. of moles of Urea = 18/60
  = 0.3 moles.

Now,
∵ Molality = No. of moles of Solute (or Urea) ÷ Mass of the Solvent in kg.
  ∴ Molality = 0.3/1.56
∴ Molality = 0.192 m.
⇒ Molality ≈ 0.2 m [Approx]

Hence Molality of the Solution is 0.192 m.(or 0.2 m.)


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When we taken mass of the urea as 15 g.
 

Given ⇒
Volume of Aqueous Solution of Urea = 15 cm³.
Density of the Solution = 1.052 g/cm³.

∵ Density = Mass/Volume
∴ Mass = Density × Volume
∴ Mass = 1.052 × 15
∴ Mass = 15.78 g.

Mass of the solution = 15.78 grams.

∵ Mass of the Solvent(or Water) = Mass of the Solution - Mass of the Solute(or Urea)
∴ Mass of the Solvent = 15.78 - 15
  = 0.78 g.
 = 78 × 10⁻⁵ kg.

Now, No. of moles of Urea = Mass/Molar Mass
 = 15/60
 = 1/4
 = 0.25 moles.


∵ Molality = No. of moles of Solute(Urea) ÷ Mass of the Solvent in kg. (Water)
  = 0.25 ÷ (78 × 10⁻⁵)
  = 320.51 m.


Hence, Molality of the solution is 320.51 m.
(It is approx impossible)

________________________


As per as my Research, First Solution is true.
In the Question, Volume of the Solution is taken as 1500 cm³.
 

Hope it helps.

Answer 2

Answer:

Explanation: see the attachment !!

Hope you easily understand ..

And i think some mistake in question is there ...of numerical value..

But my answer is absolutely correct