At 373 K,heptane and octane form an ideal so | vedclass

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(B) $1$. Calculate the moles of each component: $n_{	ext{heptane}} = \frac{25 \ g}{100 \ g \ mol^{-1}} = 0.25 \ mol$ $n_{	ext{octane}} = \frac{35 \

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$72.0$

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(B) $1$. Calculate the moles of each component: $n_{	ext{heptane}} = \frac{25 \ g}{100 \ g \ mol^{-1}} = 0.25 \ mol$ $n_{	ext{octane}} = \frac{35 \ g}{114 \ g \ mol^{-1}} \approx 0.307 \ mol$ $2$. Calculate the mole fractions: $n_{	ext{total}} = 0.25 + 0.307 = 0.557 \ mol$ $x_{	ext{heptane}} = \frac{0.25}{0.557} \approx 0.449$ $x_{	ext{octane}} = \frac{0.307}{0.557} \approx 0.551$ $3$. Apply Raoult's Law: $P_{	ext{total}} = P^{\circ}_{	ext{heptane}} \cdot x_{	ext{heptane}} + P^{\circ}_{	ext{octane}} \cdot x_{	ext{octane}}$ $P_{	ext{total}} = (105 	imes 0.449) + (45 	imes 0.551)$ $P_{	ext{total}} = 47.145 + 24.795 = 71.94 \ kPa \approx 72.0 \ kPa$.

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