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 \(= 60\, g\, mol^{-1}\))
- A \(0.027\)
- B \(0.031\)
- C \(0.028\)
- D \(0.017\)
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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