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Mila
15 days ago
10

Equimolar samples of CH3OH(l) and C2H5OH(l) are placed in separate, previously evacuated, rigid 2.0 L vessels. Each vessel is at

tached to a pressure gauge, and the temperatures are kept at 300 K. In both vessels, liquid is observed to remain present at the bottom of the container at all times. The change in pressure inside the vessel containing CH3OH(l) is shown below.Compared to the equilibrium vapor pressure of CH3OH(l) at 300 K, the equilibrium vapor pressure of C2H5OH(l) at 300 K is

Chemistry
1 answer:
Alekssandra [992]15 days ago
3 0

Answer:

Complete Question:  

Equimolar quantities of CH3OH(l) and C2H5OH(l) are placed in separate 2.0 L containers that have been evacuated beforehand. Pressure gauges are attached to each container, and the temperature is maintained at 300 K. In both containers, liquid is consistently visible at the bottom. The varying pressure within the vessel that contains CH3OH(l) is illustrated below.

In comparison to the equilibrium vapor pressure of CH3OH(l) at 300 K, the equilibrium vapor pressure of C2H5OH(l) at 300 K is

ANSWER : lower, since the London dispersion forces among C2H5OH molecules surpass those among CH3OH molecules.

Explanation:

To clarify the answer provided, let’s begin by defining some concepts.

The London dispersion force is the least strong type of intermolecular force. It is a temporary force that arises when the electron arrangement in two neighboring atoms creates transient dipoles.  

The vapor pressure of a liquid reflects the equilibrium pressure of its vapor above the liquid (or solid); specifically, it represents the pressure associated with the evaporation of a liquid (or solid) in a sealed environment above the substance.

The pressure will be lower due to the stronger London dispersion forces acting between C2H5OH molecules compared to those between CH3OH molecules. This implies that when intermolecular forces are stronger, they intensify the interactions binding the substance together, thereby reducing the liquid's vapor pressure at any given temperature and making it more difficult to vaporize the substance.

Note: The London dispersion force for C2H5OH is more substantial than for CH3OH because C2H5OH has more electrons than CH3OH.

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A chemist combined chloroform (CHCl3) and acetone (C3H6O) to create a solution where the mole fraction of chloroform is 0.187. T
KiRa [976]

Answer:

\large \boxed{\text{c = 2.50 mol/L; b = 3.96 mol/kg }}

Explanation:

1. Molar concentration

Designate chloroform as C and acetone as A.

The molar concentration for C is derived from Moles of C per Litres of solution.

(a) Moles of C

We are assuming there are 0.187 moles of C.

This resolves that step.

(b) Litres of solution

Next, identify 0.813 moles of A.

(i) Mass of each component

\text{Mass of C} = \text{0.187 mol C} \times \dfrac{\text{119.38 g C}}{\text{1 mol C}} = \text{22.32 g C}\\\\\text{Mass of A} = \text{0.813 mol A} \times \dfrac{\text{58.08 g A}}{\text{1 mol A}} = \text{47.22 g A}

(ii) Volume of each component

\text{Vol. of C} = \text{22.32 g C} \times \dfrac{\text{1 mL C}}{\text{1.48 g C}} = \text{15.08 mL C}\\\\\text{Vol. of A} = \text{47.22 g A} \times \dfrac{\text{1 mL A}}{\text{0.791 g A}} = \text{59.70 mL A}

(iii) Volume of solution

Assuming mixing doesn't alter the total volume.

V = 15.08 mL + 59.70 mL = 74.78 mL

(c) Molar concentration of C

c = \dfrac{\text{0.187 mol}}{\text{0.07478 L}} = \textbf{2.50 mol/L }\\\\\text{ The molar concentration of chloroform is $\large \boxed{\textbf{2.50 mol/L}}$}

2. Molal concentration of C

Molal concentration is calculated as moles of solute per kilograms of solvent.

Total moles of C = 0.187 mol.

Mass of A = 47.22 g = 0.047 22 kg.

\text{b} = \dfrac{\text{0.187 mol}}{\text{0.047 22 kg}} = \textbf{3.96 mol/kg }\\\\\text{The molal concentration of chloroform is $\large \boxed{\textbf{3.96 mol/kg}}$}

4 0
1 day ago
45.0 g of Ca(NO3)2 are used to create a 1.3 M solution. What is the volume of the solution
Tems11 [854]
The formula for Molarity is given by:

                                  M = moles / V
To isolate V,
                              V = moles / M ------------------(1)
Moles can also be calculated as:
                                  moles = mass / M.mass -------------(2)
Substituting the value of moles from equation 2 into equation 1 yields:
                                  V = (mass / M.mass) / M
Plugging in the numbers gives:
                                  V = (45 g / 164 g/mol) / 1.3 mol/dm³

                                       V = 0.21 dm³.
6 0
7 days ago
If the density of carbon tetrachloride is 1.59 g/ml, what is the volume in l, of 4.21 kg of carbon tetrachloride
Tems11 [854]

Density is defined as the mass-to-volume ratio. The formula for density can be expressed as:

density = \frac{mass}{volume}    -(1)

The density for carbon tetrachloride is provided as 1.59 g/ml   (given).

The mass of carbon tetrachloride is 4.21 kg   (as given).

Since, 1 kg = 1000 g

Thus, 4.21 kg = 4210 g

Utilizing the values in formula (1):

1.59 g/mL = \frac{4210 g}{volume}

volume = \frac{4210 g}{1.59 g/mL}

volume = 2647.799 mL

Since, 1 mL = 0.001 L

Hence, 2647.799 mL = 2.65 L

The resulting volume of carbon tetrachloride is 2.65 L.


6 0
11 days ago
A characteristic feature of any form of chromatography is the ________.a. calculation of an Rf value for the molecules separated
castortr0y [927]

Answer: The right choice is (c) application of both a mobile phase and a stationary phase.

Explanation:

Chromatography: This refers to a technique for separating a mixture where the mixture is distributed between two phases at varying rates, one being stationary and the other moving.

Mobile phase: The component in which the mixture is dissolved is referred to as the mobile phase.

Stationary phase: This is an adsorbent medium that remains in place while a liquid or gas passes over its surface, thus remaining stationary.

Consequently, a key characteristic of any chromatography technique involves utilizing both a mobile and a stationary phase.

4 0
2 days ago
A certain liquid has a density of 2.67 g/ cm3. what is the mass of 30.5 ml of this liquid? (
lorasvet [960]
Hello!

density = 2.67 g/cm³

volume = 30.5 mL

Thus:

Mass = density * volume

Mass = 2.67 * 30.5

Mass = 81.435 g 
4 0
7 days ago
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