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CBSE Class 12 Chemistry 2013 Solved Paper

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Question : 20 of 30
Marks: +1, -0
Determine the osmotic pressure of a solution prepared by dissolving 2.5×10−2 g2.5 \times 10^{-2}\,\mathrm{g} of K2SO4\mathrm{K}_2\mathrm{SO}_4 in 2 L2\,\mathrm{L} of water at 25∘C25^{\circ}\mathrm{C}, assuming that it is completely dissociated.
(R=0.0821 L1 atm K−1 mol−1.(R=0.0821\,\mathrm{L}_1\,\mathrm{atm}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}., Molar mass of K2SO4\mathrm{K}_2\mathrm{SO}_4 =174 g mol−1)=174\,\mathrm{g}\,\mathrm{mol}^{-1})
Solution:  
Reaction is:
K2SO4→2 K++SO42−\mathrm{K}_2\mathrm{SO}_4 \rightarrow 2\,\mathrm{K}^{+}+\mathrm{SO}_4^{2-}
No. of produced ions =3=3
π=i CRT\pi = i\,\mathrm{CRT}
π=3×0.025×0.0821×0.5×298174\pi = \frac{3 \times 0.025 \times 0.0821 \times 0.5 \times 298}{174}
=0.917174= \frac{0.917}{174}
=0.00527= 0.00527
=5.27×10−3 atm= 5.27 \times 10^{-3}\,\mathrm{atm}
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