On mixing,heptane and octane form an ideal so | vedclass

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Question

(A) Step $1$: Calculate the number of moles of each component.$n_{	ext{heptane}} = \frac{25.0 \ g}{100 \ g \ mol^{-1}} = 0.25 \ mol$$n_{	ext{octane}

Vedclass · Accepted Answer

$72.0$

Vedclass · Answer

(A) Step $1$: Calculate the number of moles of each component.$n_{	ext{heptane}} = \frac{25.0 \ g}{100 \ g \ mol^{-1}} = 0.25 \ mol$$n_{	ext{octane}} = \frac{35 \ g}{114 \ g \ mol^{-1}} \approx 0.307 \ mol$Step $2$: Calculate the mole fractions.$x_{	ext{heptane}} = \frac{0.25}{0.25 + 0.307} = \frac{0.25}{0.557} \approx 0.4488$$x_{	ext{octane}} = \frac{0.307}{0.25 + 0.307} = \frac{0.307}{0.557} \approx 0.5512$Step $3$: Apply Raoult's Law for the total vapour pressure.$P_{	ext{total}} = P_{	ext{heptane}}^{\circ} x_{	ext{heptane}} + P_{	ext{octane}}^{\circ} x_{	ext{octane}}$$P_{	ext{total}} = (105 \ kPa 	imes 0.4488) + (45 \ kPa 	imes 0.5512)$$P_{	ext{total}} = 47.124 + 24.804 = 71.928 \ kPa \approx 72.0 \ kPa$.

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