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Naily [24]
3 years ago
11

The vapor pressure of pure water at 25 °C is 23.8 torr. Determine the vapor pressure (torr) of water at 25 °C above a solution p

repared by dissolving 35 g of urea (a nonvolatile, non-electrolyte, molar mass = 60.0 g/mol) in 75 g of water.
Chemistry
1 answer:
siniylev [52]3 years ago
7 0

Answer:

Vapor pressure of solution = 20.8 Torr

Explanation:

This is about vapor pressure lowering

ΔP  = P° . Xm

ΔP states the difference between vapor pressure of pure solvent and vapor pressure from solution.

P° is the vapor pressure of pure solvent.

Xm means mole fraction (mol of solute / mol of solute + mol of solvent)

Let's determine the mol of solute

35 g / 60 g/mol = 0.583 mol of urea

Let's determine the mol of solvent

75 g / 18 g/mol = 4.167 mol of water

Total moles = 0.583 + 4.167 = 4.75

Mole fraction of solute → 0.583 /4.75 = 0.123

Let's replace the data in the formula:

23.8Torr - Vapor pressure of solution = 23.8 Torr . 0.123

Vapor pressure of solution = - (23.8 Torr . 0.123 - 23.8 Torr)

Vapor pressure of solution = 20.8 Torr

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You and your lab partner are asked to determine the density of an aluminum bar. The mass is known accurately to 4 significant fi
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Average density for method B = 2.605 g/cm³

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In order to calculate the average density for each method, we need to add the data for each method, and then divide the result by the number of measurements (in this case is 4 for both methods):

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Σ = 2.2 + 2.3 + 2.7 + 2.4 = 9.6

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C)[D]/[ED] = 5.20

D)[D]/[ED] = 5.20

E)[D']_T = 1.495* 10 ^-7 M

F)[D'] / [ED']  = 0.0579

Explanation:

E = 250 nM =2.5* 10 ^-7 mol/L , T=298.15 K

Dissociation constant of K_D = 1.3 μM (1.3 *10 ^-6 M)

E + D ⇄ ED → K_a = [ED] / [D][E]   (association constant)

ED ⇄ E + D → K_D = [E][D] / [ED]  (dissociation constant)

C)

[E] =2.5*10^-7 mol/L

K_D = 1.3* 10^-6 M

K_D = [E][D] / [ED] → [D]/[ED] = K_D / [E]

= [D]/[ED] = 1.3* 10 ^-6 / 2.5 *10^-7

= 13/25 * 10

=130/25 = 5.20

[D]/[ED] = 5.20

D)

ΔG =RTln Kd

ΔG_2 for E and D = 1.987 * 298.15 * ln 1.3*10^-6

ΔG_2 592.454 * [ln 1.3 +ln 10^-6]

ΔG_1 = 592.424 [0.2623 - 13.8155]

ΔG_2 = -592.424 * 13.553

ΔG_1 = -8184.633 cal/ mol

ΔG_1 = -8184.633  * 4.18 J/mol = -34244.508 J?mol

ΔG_1 = -34.245 KL/mol

so, ΔG_2 = ΔG_1 - 10.5 KJ/mol

ΔG_2 = -34.245 - 10.5

ΔG_2 = -44.745KJ / mol

ΔG_2 =RT ln K_D

-44.745 *10^3

=8.314 *298.15 lnK_D

lnK_D' = - 44745 / 2478.81 g

ln K_D' = -18.051

K_D' = -18.051

K_D' = e^-18.051

[D]/[ED] = 5.20

E)

[E] = 2.5* 10 ^-7 mol/ L = a

K_D' = [E][D] / [ED']                                  E +D' → ED'

K_D' = a/2(x-(a/2) / (a/2)

KD' = x - a/2

=2.447 *10^-8 = (2.5/2) * 10^-7

x=2.447 * 10^-8 + 1.25 * 10^-7

x = 2.447 *10^-8 + 1.25 * 10 ^-7

x= 10^-7 [1.25 + 0.2447]

x = 1.4947 * 10^-7

[D']_T = 1.495* 10 ^-7 M

F)

K_D' = [E][D'] / [ED']

[D'] / [ED'] = KD' / [E]

[D'] / [ED'] = 1.447 *10^-8 / 2.5* 10^-7

[D'] / [ED'] = 0.5788 * 10^-1

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