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Heptane and octane form an 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 25 g of octane ? |
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Answer» Solution :Vapour pressure of heptane `(p_(1)^(0))=105.2` kPa We know that, Molar mass of heptane `(C_(7)H_(16))` `= 7xx12+16xx1` `= 100 g mol^(-1)` Therefore, Number of moles of heptane `= (26)/(100)=0.26` mol Molar mass of octane `(C_(3)H_(18))` `= 8xx12+18xx1` = 114 g `mol^(-1)` Therefore, Number of moles of octane `= (35)/(114)=0.31` mol Mole fractionn of heptane, `x_(1)` `= (0.26)/(0.26+0.31)` = 0.456 And, mole FRACTION of oxtane, `x_(2)` `= 1-0.456` = 0.544 Now, PARTIAL pressure of heptane, `p_(1)=x_(1)p_(1)^(0)` `= 0.456xx105.2` = 47.97 kPa Partial pressure of octane, `p_(2)=x_(2)p_(2)^(0)` `= 0.544xx46.8` = 25.46 kPa Hence, vapour pressure of solution, `p_("total")=p_(1)+p_(2)` `= 47.97+25.46` = 73.43 kPa |
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