Chemistry, 05.07.2019 11:30, julia158

Amixture of 60 mol % n-propylcyclohexane and 40 mol % n-propylbenzene is distilled through a simple distillation apparatus; assume that no fractionation occurs during the distillation. the boiling temperature is found to be 157 degrees celsius (760 torr) as the first small amount of distillate is collected. the standard vapor pressures of n-propylcyclohexane and n-propyl-benzene are known to be 769 torr and 725 torr, respectively, at 1567.3 degrees celsius. calculate the percentage of each of the two components in the first few drops of distillate.

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A mixture of 60 mol % n-propyl cyclohexane and 40 mol % n-propylbenzene is distilled through a simple distillation apparatus; assume that no fractionation occurs during the distillation, the percentage of each of the two components in the first few drops of distillate are 0.61 and 0.39 respectively.

From the information given, the first thing to do is to determine the mole fraction of each substance.

$\mathbf{The \ mole \ fraction = \dfrac{mole \ percent }{100}}$

For n-propyl cyclohexane;

$\mathbf{The \ mole \ fraction \ is = \dfrac{60 }{100}}$

The mole fraction is = 0.60 mole

For n-propylbenzene;

$\mathbf{The \ mole \ fraction \ is = \dfrac{40 }{100}}$

The mole fraction is = 0.40 mole

By the use of Raoult's law, we can estimate the total vapor pressure of the solution as follows:

$\mathbf{P{total} = P^0_{cy} N_{cy} + P^0_{benz}N_{benz}}$

where;

$\mathbf{P^0_{cy} }$ = n-propylcyclohexane vapor pressure $\mathbf{N_{cy} }$ = n-propylcyclohexane mole fraction $\mathbf{P^0_{benz} }$ = n-propylbenzene vapor pressure $\mathbf{N_{benz} }$ = n-propylbenzene mole fraction

Replacing the values given in the question

$\mathbf{P{total} = (769 \ torr\times 0.6) +(725\ torr \times 0.4)}$

$\mathbf{P{total} = (461 \ torr) +(290 \ torr)}$

$\mathbf{P{total} = 751.4 \ torr}$

The percentage composition of each component can now be estimated as follows:

The percentage composition can be determined by finding the ratio of the vapor pressure of the component given by the total vapor pressure.

Percentage component of n-propyl cyclohexane is;

= $\mathbf{\dfrac{461.4}{751.4}}$

= 0.61

Percentage component of n-propylbenzene is;

= $\mathbf{\dfrac{290}{751.4}}$

= 0.39

Therefore, we can conclude that the percentage of each of the two components in the first few drops of distillate is 0.61 for n-propyl cyclohexane and 0.39 for n-propylbenzene

Learn more about percentage composition here:

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answered: brinleychristofferse

61 mole % propylcyclohexane and 39 mole % propylbenzene

Explanation:

For convenience, let’s call propylcyclohexane Component 1 and propylbenzene Component 2.

According to Raoult’s Law,

$p_{1} = \chi_{1}p_{1}^{\circ}$ and

$p_{2} = \chi_{2}p_{2}^{\circ}$

where

p₁ and p₂ are the vapour pressures of the components above the solution

χ₁ and χ₂ are the mole \fractions of the components

p₁° and p₂° are the vapour pressures of the pure components.

So,

$p_{1} = 0.60 \times \text{769 torr} = \text{ 461 torr}$

$p_{2} = 0.40 \times \text{725 torr} = \text{ 290 torr}$

$p_{\text{tot}}$ = p₁ + p₂= 461 torr + 290 torr = 751 torr

∴ In the vapour

$\chi_{1} = \frac{p_{1}}{p_{\text{tot}} } = \frac{\text{461 torr} }{\text{751 torr}} = 0.61$

χ₂ = 1 – χ₁ = 1 - 0.61 = 0.39

answered: Guest

most likely he is cheating on you

answered: Guest

Alpha particles have positive charge, beta particles have negative charge, and answer 4 just doesn’t make sense. thus, you are left with only one possible answer.