Test Index

NCERT Class XII Chemistry
Chapter - Solutions
Questions with Solutions

© examsnet.com
Question : 50 of 53
Marks: +1, -0
Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea NH2CONH2\mathrm{NH_2CONH_2} is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
Solution:  
Given, P0=23.8 mm HgP^{0}=23.8\,\text{mm}\,\text{Hg}
w2=50 g,M2(w_{2}=50\,\text{g}, M_{2}( urea )=60 g mol−1,)=60\,\text{g}\,\text{mol}^{-1},
Ps=?P0−PsP0=?P_{s}=? \frac{P^{0}-P_{s}}{P^{0}}=?
w1=850 g,M1(w_{1}=850\,\text{g}, M_{1}( water )=18 g mol−1)=18\,\text{g}\,\text{mol}^{-1}
∴  n2=5060=0.83\therefore\; n_{2}=\frac{50}{60}=0.83
∴n1=85018=47.22\therefore n_{1}=\frac{850}{18}=47.22
Applying Raoult's law, P0−PsP0=n2n1+n2\frac{P^{0}-P_{s}}{P^{0}}=\frac{n_{2}}{n_{1}+n_{2}}
or,P0−PsP0=0.8347.22+0.83\frac{P^{0}-P_{s}}{P^{0}}=\frac{0.83}{47.22+0.83}
=0.8348.05=0.017=\frac{0.83}{48.05}=0.017
Thus, relative lowering of vapour pressure = 0.017
Again, ΔPP0=0.017\frac{\Delta P}{P^{0}}=0.017
⇒ΔP=0.017×23.8\Rightarrow \Delta P=0.017 \times 23.8
or, P0−Ps=0.017×23.8P^{0}-P_{s}=0.017 \times 23.8
or, Ps=23.8−0.017×23.8P_{s}=23.8-0.017 \times 23.8
or Ps=23.4 mm HgP_{s}=23.4\,\text{mm}\,\text{Hg}
Thus, vapour pressure of water in the solution = 23.4 mm Hg
© examsnet.com
Go to Question:
Prev Question
Next Question