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

linearly: 8560 bacteria

exponentially: approximately 75233 bacteria

step-by-step explanation:

assuming it is not exponential (we would have some really complicated logarithms), let's say it is linear. then, after 5 hours, it increased by 1200 bacteria, so every hour it increases by 240 bacteria. this means that after 19 hours, we get 19 * 240 + 4000 bacteria or 3800 + 760 + 4000 = 7800 + 760 = 8560 bacteria.

because it says round your answer, i will do this exponentially (with a calculator, of course). first, we need to find how much it multiplies by every hour, by taking the 5th root of 52/40 or 13/10. this is approximately 0.167. we still need to add 1 to this, because we are multiplying by that number, not adding that amount of the number. now, we just calculate 4000 * (1.167)^19. use a calculator to get approximately 75232.6, or 75233.

Answer from: Quest

Im sorry but what on

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