Biphenyl, C12H10, is a nonvolatile, nonionizing solute that is soluble in benzene, C6H6. At 25 °C, the vapor pressure of pure benzene is 100.84 torr. What is the vapor pressure of a solution made from dissolving 15.4 g of biphenyl in 27.2 g of benzene?
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first of all since biphenyl is non ionising, take vant hoff factor to be 1. Now calculate mole fraction of biphenyl in benzene. Let moles of the two be x and y respectively. Then x = 15.4/molar mass of biphenyl. And y= 27.2/molar mass of benzene. Now mole fraction of biphenyl in benzene is x/(x+y). Suppose this value comes out to be z.
Now let the vapour pressure of pure benzene be P. Let vapour pressure of solution be P'. Hence-
P'=zP
You can calculate P' using this relation :)
Now let the vapour pressure of pure benzene be P. Let vapour pressure of solution be P'. Hence-
P'=zP
You can calculate P' using this relation :)