On mixing, heptane and octane form an ideal solution At 373 K, the vapour pressures of the two liquidcomponents (Heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of the solutionobtained by mixing 25.0 g of heptane and 35 g of octane will be (molar mass of heptane = 100 g 1 mol-and of octane =114 g 1 mol- )

let the partial pressure of heptane and octane of the solution be ph and poct
now partial pressure of any component in solution is given by=pressure in pure solution*mole fraction----(1)
mole of heptane=25/100=.25 and mole of octane=35/114=.3
now mole fraction of heptane=.25/.55=.45
mole fraction of octane=.3/.55=.54
now partial pressure of heptane=.45*105=47.25KPa
partial pressure of octane=45*.54=24.3Kpa
thus the total pressure is the sum of the individual pressure in the solution=47.2+24.3Kpa=70.2Kpa