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NCERT Class XII Chemistry
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Question : 18 of 53
Marks: +1, -0
Calculate the mass of a non-volatile solute (molar mass 40 g mol−1^{-1}) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
Solution:  
Number of moles of solute =w40 mol= \frac{w}{40}\,\mathrm{mol}
Let P0=100P^{0}=100 then Ps=80,ΔP=20P_{s}=80, \Delta P=20
No. of moles of solvent (octane) =114 g114 g mol−1=1 mol= \frac{114\,\mathrm{g}}{114\,\mathrm{g}\,\mathrm{mol}^{-1}} = 1\,\mathrm{mol}
[molar mass of C8H18=114 g mol−1]\mathrm{C}_8\mathrm{H}_{18}=114\,\mathrm{g}\,\mathrm{mol}^{-1} ]
Now, ΔPP0=x2  ∴  20100=w40w40+1\frac{\Delta P}{P^{0}} = x_{2} \;\therefore\; \frac{20}{100} = \frac{\frac{w}{40}}{\frac{w}{40}+1}
or, 0.2(w40+1)=w400.2\left(\frac{w}{40}+1\right) = \frac{w}{40} or, 0.8w40=0.2\frac{0.8w}{40}=0.2 or w=10 gw=10\,\mathrm{g}
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