Question

Heptane and octane form 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 35.0 g of octane?​

Answer 1

Answer :

• Vapour pressure of the mixture = 73.43 kPa.

Step-by-step explanation :

Given that,

• Vapour pressure of heptane = 105.2 kPa.
• Vapour pressure of octane = 46.8 kPa.

Also,

• Molar mass of heptane = 100 g/mol.
• Molar mass of octane = 114 g/mol.

$\hrulefill$

$\rm{No.\:of\:moles\:of\:heptane=\dfrac{26.0\:g}{100\:g\:mol^{-1}}=0.26\:mol}$

$\rm{No.\:of\:moles\:of\:octane=\dfrac{35.0\:g}{114\:g\:mol^{-1}}=0.31\:mol}$

$\rm{Now,\:x(heptane)=\dfrac{0.26\:g}{0.26+0.31}=0.456}$

$\rm{x(octane)=1-0.456=0.544}$

$\rule{200}{3}$

$\rm{Now,\:p(heptane)=0.456\times{}105.2\:kPa=47.97\:kPa}$

$\rm{p(octane)=0.544\times{}46.8\:kPa=25.46\:kPa}$

$\therefore\rm{P_{total}=47.97+25.46=}\bold{73.43\:kPa.}$