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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 \

Vedclass · Accepted Answer

$72.0$

Vedclass · Answer

(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: $P_{	ext{total}} = P^{\circ}_{	ext{heptane}} \cdot x_{	ext{heptane}} + P^{\circ}_{	ext{octane}} \cdot x_{	ext{octane}}$ $P_{	ext{total}} = (105 	imes 0.449) + (45 	imes 0.551)$ $P_{	ext{total}} = 47.145 + 24.795 = 71.94 \ kPa \approx 72.0 \ kPa$.

Similar Questions

When $25 \ mL$ of acetone is mixed with $25 \ mL$ of ethanol,the volume of the resulting solution will be ...........

Which one of the following non-ideal solutions shows negative deviation?

Which of the following liquid mixtures shows a positive deviation from Raoult's law?

For an ideal solution of two components $A$ and $B$,which of the following is true?

$p_A$ and $p_B$ are the vapour pressure of pure liquid components,$A$ and $B,$ respectively of an ideal binary solution. If $x_A$ represents the mole fraction of component $A,$ the total pressure of the solution will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module