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# 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?

2.16     Heptane and octane form an ideal solution. At $\inline 373\; K$, the vapour pressures of the two liquid components are $\inline 105.2\; kPa$ and $\inline 46.8\; kPa$  respectively. What will be the vapour pressure of a mixture of $\inline 26.0\; g$ of heptane and $\inline 35\; g$  of octane?

Views

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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