The boiling point of an equimolar ideal solut | vedclass

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(B) For an ideal solution,the boiling point $(T_b)$ depends on the mole fraction $(x)$ of the components. Let the mass of each component be $m$. Moles

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Greater than $80 \ ^oC$

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(B) For an ideal solution,the boiling point $(T_b)$ depends on the mole fraction $(x)$ of the components. Let the mass of each component be $m$. Moles of hexane $(n_1)$ = $m / 86$ and moles of heptane $(n_2)$ = $m / 100$. Mole fraction of hexane $(x_1)$ = $n_1 / (n_1 + n_2) = (m/86) / (m/86 + m/100) = 100 / 186 \approx 0.537$. Mole fraction of heptane $(x_2)$ = $1 - 0.537 = 0.463$. The boiling point of the mixture is given by $T_b = x_1 T_{b1} + x_2 T_{b2}$ (approximately for ideal mixtures). $T_b = (0.537 	imes 69) + (0.463 	imes 98) \approx 37.05 + 45.37 = 82.42 \ ^oC$. Since the mole fraction of the lower boiling component (hexane) is higher in the equal mass mixture $(0.537)$ compared to the equimolar mixture $(0.5)$,the boiling point will be slightly higher than the equimolar boiling point of $80 \ ^oC$ but less than $98 \ ^oC$. Thus,the boiling point is between $80 \ ^oC$ and $98 \ ^oC$.

The boiling point of an equimolar ideal solution of hexane $(C_6H_{14})$ and heptane $(C_7H_{16})$ is $80 \ ^oC$. What is the boiling point of a mixture of hexane and heptane containing equal mass of both? The boiling points of pure hexane and pure heptane are $69 \ ^oC$ and $98 \ ^oC$ respectively.

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