Mathematics, 03.03.2020 22:42 upadrastakameswari

Pentane (C5H12) and hexane (C6H14) combine to form an ideal solution. At 258C the vapor pressures of pen- tane and hexane are 511 and 150. torr, respectively. A solution is prepared by mixing 25 mL of pentane (density 5 0.63 g/mL) with 45 mL of hexane (density 5 0.66 g/mL).
(a) What is the vapor pressure of this solution? torr
(b) What is the mole fraction of pentane in the vapor that is in equilibrium with this solution?

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Answer from: Jadiahd

Step-by-step explanation:

(a) The given data is as follows.

$P^{o}_{1}$ = 511 torr, $P^{o}_{2}$ = 150 torr

Mass of pentane = volume × density

= $25 ml \times 50.63 g/mL$

= 1265.75 g

Mass of hexane = volume × density

= $45 ml \times 50.66 g/ml$

= 2279.7 g

Now, we will calculate the moles of both pentane and hexane as follows.

No. of moles of pentane = $\frac{mass}{\text{molar mass}}$

= $\frac{1265.75 g}{72 g/mol}$

= 17.58 mol

No. of moles of hexane = $\frac{mass}{\text{molar mass}}$

= $\frac{2279.7 g}{86 g/mol}$

= 26.51 mol

Tota number of moles = 17.58 mol + 26.51 mol

= 44.09 mol

Mole fraction of pentane ( $X_{1}$ ) = $\frac{moles}{\text{total moles}}$

= $\frac{17.58 mol}{44.09 mol}$

= 0.39

Mole fraction of hexane ( $X_{2}$ ) = $\frac{moles}{\text{total moles}}$

= $\frac{26.51 mol}{44.09 mol}$

= 0.60

So, we will calculate the vapor pressure of the solution as follows.

$P_{solution} = x_{1}P^{o}_{1} + x_{2}P^{o}_{2}$

where, $x_{1}$ = mole fraction of solution 1

$P^{o}_{1}$ = partial pressure of pure solvent of solution 1

$x_{2}$ = mole fraction of solution 2

$P^{o}_{2}$ = partial pressure of pure solvent of solution 2

$P_{solution} = x_{1}P^{o}_{1} + x_{2}P^{o}_{2}$

= $0.39 \times 511 + 0.60 \times 150$

= (199.29 + 90) torr

= 289.29 torr

Therefore, vapor pressure of this solution is 289.29 torr.

(b) Mole fraction of pentane in the vapor that is in equilibrium with this solution is calculated as follows.

= $\frac{\text{Partial pressure of pentane}}{\text{total pressure}}$

= $\frac{0.39}{289.29}$

= $1.34 \times 10^{-3}$

Hence, mole fraction of pentane in the vapor that is in equilibrium with this solution is $1.34 \times 10^{-3}$ .

Answer from: Quest

step-by-step explanation:

12: 00 p.m. to 7: 00 p.m. is a total of 7 hours.

4: 00 p.m. to 7: 00 p.m. is a total of 3 hours.

this means that from 4: 00 p.m to 7: 00 p.m. this represents 3/7 of what was lost from 12: 00 p.m. to 7: 00 p.m.

after knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

we then divide 4,800 by 3 which gets 1,600.

because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200. after adding 11,200 to how much water was left at 7: 00 p.m. which was 6,800. 11,200 + 6,800 = 18,000

this means that the pool originally had 18,000 gallons in the pool.

you can also check your work by doing 18,000 - 4(1600), which becomes 1800 - 6400 which gets you 11,600.

Answer from: Quest

i think asking google will you get the answer to that question

Find the center of the circle whose equation is x2 + y2 =4

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Pentane (C5H12) and hexane (C6H14) combine to form an ideal solution. At 258C the vapor pressures of...

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