Question

Heptane and octane form an ideal solution. At 373 K, the vapour pressure 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?

Answer 1

Hint: According to Raoult’s law, for a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in the solution.
Thus, $$P_1\propto X_1$$
Or,$$P_1=P_1^0\times X_1$$, where
$$P_1=Partialvapourpressureof\text{ heptane}$$
$$P_1^0=Purevapourpressureof\text{ heptane}$$
$$X_1=Molefractionof\text{ heptane}$$

Similarly,$$P_2=P_2^0\times X_2$$,
where
$$P_2=Partialvapourpressureof\text{ octane}$$
$$P_2^0=Purevapourpressureof\text{ octane}$$
$$X_2=Molefractionof\text{ octane}$$
According to Dalton’s law of partial pressure, the total pressure of the mixture will be the sum of the partial pressure of each component of the solution.
$$P_{total}=P_1+P_2$$

Complete step by step answer:
Here it is given that,
Vapour pressure of heptane in pure form ($$P_1^0$$) = 105.2 kPa
Vapour pressure of octane in pure form ($$P_2^0$$) = 46.8 kPa
We will calculate the moles of heptane and octane to find their mole fraction.
Molar mass of heptane ($$C_7{H}_{16}$$) = 12 x 7 + 16 x 1 = 100 g/mol

Number of moles = ${\displaystyle \frac{Givenmass}{Molarmass}}$

$$Molesofheptane={\displaystyle \frac{26\text{ g}}{100\text{ g/mol}}}$$ = 0.26 mol
Molar mass of octane ($$C_8{H}_{18}$$) = 12 x 8 + 18 x 1 = 114 g/mol

$$Molesofoctane={\displaystyle \frac{35\text{ g}}{114\text{ g/mol}}}$$ = 0.30 mol

Total number of moles = 0.26 + 0.30 = 0.56 mol
$$Therefore\text{Mole fraction of heptane }\left(X_{C_7{H}_{16}}\right)\text{ = }{\displaystyle \frac{\text{0}\text{.26 }}{0.56}}$$= 0.46
And $$\text{Mole fraction of octane }\left(X_{C_8{H}_{18}}\right)\text{ = }{\displaystyle \frac{\text{0}\text{.30 }}{0.56}}$$ = 0.54
According to Dalton’s law, $$P_{total}=P_1+P_2$$
Or, $$P_{total}=P_1^0\times X_1+P_2^0\times X_2$$ (According to Raoult’s law $$P_1=P_1^0\times X_1$$ and $$P_2=P_2^0\times X_2$$)
Or, $$P_{total}=P_1^0{X}_{C_7{H}_{16}}+P_2^0{X}_{C_8{H}_{18}}$$
Now we will put the values of variables in above equation,
$$P_{total}=105.2\text{ kPa }\times \text{ 0}\text{.46 + 46}\text{.8 kPa }\times \text{ 0}\text{.54}$$
$$P_{total}=\text{ 73}\text{.66 kPa}$$
Hence the vapour pressure of mixture containing 26 g of heptane and 35 g of octane is 73.66 kPa.

Note: To solve this question we should know to calculate the molecular masses of heptane and octane and the formula for calculating their mole fraction. Some time should make mistakes by getting confused in partial pressure and total pressure but one should remember that in Raoult’s law and Dalton’s law partial pressure is used and then after getting the values of partial pressure we can further calculate total pressure