2.16     Heptane and octane form an ideal solution. At 373\; K, the vapour pressures 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\; g  of octane?

Answers (1)

Vapour pressure of heptane =   p_h^{\circ} = 105.2\ KPa 

and vapour pressure of octane =  p_o^{\circ} = 46.8\ KPa 

Firstly we will find moles of heptane and octane so that we can find vapour pressure of each.

Molar mass of heptane =  7(12) + 16(1)  =  100 unit.

and molar mass of octane =  8(12) + 18(1) = 114 unit.

So moles of heptane :

                                         \frac{26}{100} = 0.26

and moles of octane :

                                        \frac{35}{114} = 0.31

Mole fraction of heptane =  0.456 and mole fraction of octane = 0.544 

Now we will find the partial vapour pressure:-

                      (i) of heptane :-  p_h = 0.456\times105.2 = 47.97\ KPa            

                      (ii) of octane  :-   p_o = 0.544\times46.8 = 25.46\ KPa

So total pressure of solution = p_h+p_o

                                             =  47.97 + 25.46 =   73.43 KPa

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