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NCERT Class XII Chemistry
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Question : 53 of 53
Marks: +1, -0
Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C.
Solution:  
Given, Number of moles of solute dissolved (n)
=1.0 g185,000 g mol−1=\frac{1.0\,\mathrm{g}}{185,000\,\mathrm{g}\,\mathrm{mol}^{-1}}
=1185,000 mol=\frac{1}{185,000}\,\mathrm{mol}
V=450 mL=0.450 L,V=450\,\mathrm{mL}=0.450\,\mathrm{L},
T=37∘ CT=37^{\circ}\,\mathrm{C}
=37+273=310 K=37+273=310\,\mathrm{K} RR =8.314 kPa L K−1 mol−1=8.314\,\mathrm{kPa}\,\mathrm{L}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}
=8.314×103 Pa L K−1 mol−1=8.314\times10^{3}\,\mathrm{Pa}\,\mathrm{L}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}
Ï€=CRT=nVRT\pi=C R T=\frac{n}{V} R T
Substituting these values, we get
π=\pi= 1185,000×10.45×8.314×103×310\frac{1}{185,000}\times\frac{1}{0.45}\times8.314\times10^{3}\times310
=30.96 Pa=30.96\,\mathrm{Pa}
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