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Vapour pressure of chloroform (CHCl_(3)) and dichloromethane (CH_(2) Cl_(2)) at 298 K and 200 mm Hg and 415 mm Hg respectively (i) calculate the vapour pressure of the solution prepared by mixing 25·5 g of CHCl_(3) and 40 g ef CH_(2)Cl_(2) at 298 K and (ii) mole fraction of each component in vapour phase. |
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Answer» Solution :(i) MOLAR mass of`CH_(2)Cl_(2)` `(12 xx 1 + 1 xx 2 + 35.5 xx 2 )` = 85 g` mol^(-1)` Molar mass of `CHCl_(3)` = `(12 xx 1 + 1 xx 1 + 35.5 xx 3) ` `= 119.5 g mol^(-1)` moles of `CH_(2)Cl_(2) = (40)/(85)` = 0.47 mol Moles of `CHCl_(3) = (25.5)/(119.5) = 0.213 ` mol Total number of moles in the solution =0·47 + 0·213 = 0.683 mol `X CH_(2) Cl_(2) = ("mole of "CH_(2) Cl_(2))/("Total mole of solutions")` `= (0.47)/(0.683) = 0.688` `X CH_(2) Cl_(2) = (1.00 - 0.688) = 0.312` Now we know that ` P_("total") = P_(A)^(0) + (P_(A)^(0) - P_(A)^(0) ) xx CH_(2) Cl_(2)` `X CH_(2) Cl_(2) = 200 + (415 - 200) xx 0.688` = 200 + 147.9 = 347.9 mm Hg (ii) Mole fraction of component in vapour phase `P_(A)^(0) = CHCl_(3) ` `P_(B)^(0) = CH_(2) Cl_(2)` `P_(CH_(2)Cl_(2)) = P_(CH_(2) xx Cl_(2))^(0) xx X_(CH_(2)Cl_(2))` = 415 `xx `0.688 = 285.5 mm Hg `P_(CHCl_(3)) = P_(CHCl_(3))^(0) xx X_(CHCl_(3))` = 200 `xx 0.312` = 62.4 mm Hg P = y `xx P_("total")` `Y_(CH_(2)Cl_(2)) = (P_(CH_(4)Cl_(2))/(P_("total")))` ` = (285.5)/(347.9) = 0.82` `Y_(CHCl_(3)) = (P_(CHCl_(3))/(P_("Total")))` `= (62.4)/(347.9) = 0.18` |
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