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NCERT Class XII Chemistry
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Question : 16 of 53
Marks: +1, -0
Heptane and octane form ideal solution. At 373 K, the vapour pressure of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35.0 g of octane?
Solution:  
Molar mass of heptane C7H16=100 g mol−1\mathrm{C}_7\mathrm{H}_{16} = 100 \, \mathrm{g} \, \mathrm{mol}^{-1}
Molar mass of octane C8H18=114 g mol−1\mathrm{C}_8\mathrm{H}_{18} = 114 \, \mathrm{g} \, \mathrm{mol}^{-1}
Number of moles of heptane =26.0 g100 g mol−1=0.26 mol= \frac{26.0 \, \mathrm{g}}{100 \, \mathrm{g} \, \mathrm{mol}^{-1}} = 0.26 \, \mathrm{mol}
No. of moles of heptane =35.0 g114 g mol−1=0.31 mol= \frac{35.0 \, \mathrm{g}}{114 \, \mathrm{g} \, \mathrm{mol}^{-1}} = 0.31 \, \mathrm{mol}
xx (heptane )=0.260.26+0.31=0.456= \frac{0.26}{0.26+0.31} = 0.456
xx (octane )=1−0.456=0.544) = 1 - 0.456 = 0.544
xx (octane )=1−0.456=0.544) = 1 - 0.456 = 0.544
p(p( heptane =0.456×105.2 kPa=47.97 kPa= 0.456 \times 105.2 \, \mathrm{kPa} = 47.97 \, \mathrm{kPa}
p(p( octane =0.544×46.8 kPa=25.46 kPa= 0.544 \times 46.8 \, \mathrm{kPa} = 25.46 \, \mathrm{kPa}
PTotal=47.97+25.46=73.43 kPaP_{\text{Total}} = 47.97 + 25.46 = 73.43 \, \mathrm{kPa}
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