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JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties

At room temperature, a dilute solution of urea is prepared by dissolving \(0.60\, g\) of urea in \(360\, g\) of water. If the vapour pressure of pure water at this temperature is \(35\, mm\, Hg\), lowering of vapour pressure will be: .............. \(\mathrm{mm\,Hg}\)  (molar mass of urea \(= 60\, g\, mol^{-1}\))

  1. A \(0.027\)
  2. B \(0.031\)
  3. C \(0.028\)
  4. D \(0.017\)
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Answer & Solution

Correct Answer

(D) \(0.017\)

Step-by-step Solution

Detailed explanation

Lowering of vapour pressure \( = \,{p^o}\, - \,p\, = \,{p^o}.{X_{solute}}\) \(\therefore \,\,\Delta p\, = \,35\, \times \,\frac{{0.6/60}}{{\frac{{0.6}}{{60}} + \frac{{360}}{{18}}}}\) \( = 35\, \times \,\frac{{0.01}}{{0.01 + 20}}\, = 35\, \times \,\frac{{0.1}}{{20.01}}\)…
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