The density of toluene (C7H8) is 0.867 g/mL, and the density of thiophene (C4H4S) is 1.065 g/mL. A solution is made by dissolving 8.10 g of thiophene in 250.0 mL of toluene.
(a) Calculate the mole fraction of thiophene in the solution.
(b) Calculate the molality of thiophene in the solution.
(c) Assuming that the volumes of the solute and solvent are additive, what is the molarity of thiophene in the solution?

Answers

Answer 1

Considering the solution of mole fraction, molality and molarity, you obtain that:

(a) the mole fraction of thiophene in the solution is 0.039.

(b) the molality of thiophene in the solution 0.4429 [tex]\frac{moles}{kg}[/tex].

(c)  the molarity of thiophene in the solution is 0.373[tex]\frac{moles}{L}[/tex].

You know that:

Density toluene (C₇H₈)= 0.867 [tex]\frac{g}{mL}[/tex]Density thiophene (C₄H₄S)= 1.065 [tex]\frac{g}{mL}[/tex] Volume of toluene (C₇H₈)= 250 mL= 0.250 L (being 1000 mL= 1 L)Mass of thiophene (C₄H₄S)= 8.10 grams

(a) Mole fraction

The molar fraction is a way of measuring the concentration that expresses the proportion in which a substance is found with respect to the total moles of the solution.

Being the molar mass of each compound equals to:

Toluene (C₇H₈)= 92 [tex]\frac{g}{mol}[/tex]Thiophene (C₄H₄S)= 84 [tex]\frac{g}{mol}[/tex]

the number of moles of each compound can be calculated as:

Toluene (C₇H₈)= [tex]250 mLx\frac{0.867 grams}{1 mL} x \frac{1 mole}{92 grams}[/tex]= 2.35 molesThiophene (C₄H₄S)= [tex]8.10 gramsx\frac{1 mole}{84 grams}[/tex]= 0.096 moles

The total moles is obtained from the addition of the moles of the solute (C₄H₄S) and the solvent (C₇H₈):

total moles = moles C₄H₄S + moles C₇H₈ = 0.096 moles + 2.35 moles = 2.45 moles

Then, the mole fraction of thiophene in the solution can be calculated as:

[tex]mole fraction of thiophene=\frac{0.096 moles}{2.45 moles}[/tex]

Solving:

mole fraction of thiophene= 0.039

Finally, the mole fraction of thiophene in the solution is 0.039.

(b) Molality

Molality is the ratio of the number of moles of any dissolved solute to kilograms of solvent.

The Molality of a solution is determined by the expression:

[tex]Molality=\frac{number of moles of solute}{kilograms of solvent}[/tex]

In this case, you know:

number of moles of solute (C₄H₄S)= 0.096 moles Mass of solvent = [tex]250 mLx\frac{0.867 grams}{1 mL}[/tex] = 216.75 g = 0.21675 kg (being 1000 g=1 kg)  

Replacing:

[tex]Molality C_{4} H_{4}S =\frac{0.096 moles}{0.21675 kg}[/tex]

molality C₄H₄S= 0.4429 [tex]\frac{moles}{kg}[/tex]

Finally, the molality of thiophene in the solution 0.4429 [tex]\frac{moles}{kg}[/tex].

(c) Molarity

Molarity is the number of moles of solute that are dissolved in a certain volume and is determined by the following expression:

[tex]Molarity=\frac{number of moles of solute}{volume}[/tex]

Assuming that the volumes of solute and solvent are additive, you can add the volume of C₄H₄S and C₇H₈.

But first, you need yo know the volume of C₄H₄S, which can be calculated from the mass and density:

[tex]Volume C_{4} H_{4}S =8.10 grams\frac{1 mL}{1.065 grams}[/tex]= 7.606 mL= 0.007606 L

Then, the total volume of the solution is calculated as:

total volume of the solution= volume C₇H₈ + volume C₄H₄S

total volume of the solution= 0.250 L + 0.007606 L = 0.257606 L

So, the molarity of thiophene in the solution can be calculated as:

[tex]Molarity C_{4} H_{4} S=\frac{number of moles of C_{4} H_{4} S}{totalvolumeof the solution}[/tex]

Replacing:

[tex]Molarity C_{4} H_{4} S=\frac{0.096 moles}{0.257606 L}[/tex]

Solving:

Molarity C₄H₄S= 0.373 [tex]\frac{moles}{L}[/tex]

Finally, the molarity of thiophene in the solution is 0.373[tex]\frac{moles}{L}[/tex].

In summary, you get:

(a) the mole fraction of thiophene in the solution is 0.039.

(b) the molality of thiophene in the solution 0.4429 [tex]\frac{moles}{kg}[/tex].

(c)  the molarity of thiophene in the solution is 0.373[tex]\frac{moles}{L}[/tex].

Learn more about

density: brainly.com/question/952755?referrer=searchResults brainly.com/question/1462554?referrer=searchResults mole fraction: brainly.com/question/14434096?referrer=searchResults brainly.com/question/10095502?referrer=searchResults molality brainly.com/question/20366625?referrer=searchResults brainly.com/question/4580605?referrer=searchResults molarity with this example: brainly.com/question/15406534?referrer=searchResults

Related Questions

True or false: potential energy increases as like charges get closer to one another

Answers

True, is the correct answer.

Assuming that only the listed gases are present, what would be the mole fraction of oxygen gas be for each of the following situations?

A gas sample of 2.31 atm of oxygen gas and 3.75 atm of hydrogen gas that react to form water vapor. Assume the volume of the container and the temperature inside the container does not change.​

Answers

The mole fraction of oxygen gas : 0.381

Further explanation

Given

2.31 atm Oxygen

3.75 atm Hydrogen

Required

The mole fraction of Oxygen

Solution

Dalton Law's of partial pressure

P tot = P₁ + P₂ + ..Pₙ

Input the value :

P tot = P O₂ + P H₂

P tot = 2.31 atm + 3.75 atm

P tot = 6.06 atm

The mole fraction of O₂ (X O₂):

P O₂ = X O₂ x P tot

X O₂ = P O₂ / P tot

X O₂ = 2.31 /6.06

X O₂ = 0.381

The mole fraction of oxygen gas in the given mixture of oxygen and hydrogen gas is 0.381.

How do we calculate mole fraction?

If the details is present in terms of partial pressure then by using the below equation, we can calculate the mole fraction as:

p = (x)(P), where

p = partial pressure

P = total pressure

x = mole fraction

Total pressure of the mixture will be calculated by adding the partial pressure of oxygen and hydrogen gases as:

P = 2.31 + 3.75 = 6.06 atm

Now mole fraction of oxygen will be calculated as:

x = 2.31 / 6.06 = 0.381

Hence mole fraction of oxygen is 0.381.

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Answers

Answer:

Calcium

Explanation:

The symbol for the element with a mass number of 27 is actually Al for Aluminum instead of Co for Cobalt. Mass number refers to atomic mass, not atomic number.

Aluminum has an atomic mass of 26.982 or 27.

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The mystery element is Calcium.

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How many grams of iron(III) hydroxide (106.87 g/mol) will precipitate if 50.0 mL of 0.500 M sodium hydroxide is added to 75.0 mL of 0.200 M iron(III) nitrate

Answers

Answer:

[tex]m_{Fe(OH)_3} = 0.891gFe(OH)_3[/tex]

Explanation:

Hello!

In this case, given the chemical reaction:

[tex]3NaOH(aq)+Fe(NO_3)_3(aq)\rightarrow 3NaNO_3(aq)+Fe(OH)_3(s)[/tex]

In such a way, given the volumes and molarities of each reactant, we can compute the moles of produced iron (III) hydroxide by each of them, via the 3:1 and 1:1 mole ratios:

[tex]n_{Fe(OH)_3}^{by\ NaOH}=0.0500L*0.500\frac{molNaOH}{L}*\frac{1molFe(OH)_3}{3molNaOH} =0.00833molFe(OH)_3\\\\n_{Fe(OH)_3}^{by\ Fe(NO_3)_3}=0.0750L*0.200\frac{molFe(NO_3)_3}{L}*\frac{1molFe(OH)_3}{1molFe(NO_3)_3} =0.0150molFe(OH)_3[/tex]

It means that the sodium hydroxide is the limiting reactant and 0.00833 moles of iron (III) hydroxide are produced; thus, the required mass is:

[tex]m_{Fe(OH)_3}=0.00833molFe(OH)_3*\frac{106.87gFe(OH)_3}{1molFe(OH)_3} \\\\m_{Fe(OH)_3} = 0.891gFe(OH)_3[/tex]

Best regards!

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KD 0.18 moles
1.3 moles
11 moles

Answers

Answer:

0.089

Explanation:

2.5/ 14= .178 then 0.178/ 2 again and you get 0.089

The moles of nitrogen in 2.5 grams of the compound is 0.089 moles. Thus option A is correct.

Moles can be calculated as the ratio of mass to molecular mass.

Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]

The mass of Nitrogen = 14 g/mol

The mass of diatomic nitrogen = 2 [tex]\times[/tex] mass of nitrogen

The mass of diatomic nitrogen = 2 [tex]\times[/tex] 14 g/mol

The mass of diatomic nitogen = 28 g/mol

The molecular weight of Nitrogen = 28g/mol

Given, the mass of Nitrogen = 2.5 g

Moles of nitrogen = [tex]\rm \dfrac{2.5}{28}[/tex]

Moles of Nitrogen = 0.089 moles.

The moles of nitrogen in 2.5 grams of the compound is 0.089 moles. Thus option A is correct.

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Explanation:

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The scientist Alfred Wegener.

Explanation:

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C+02 = CO2

Answers

[tex]\tt Kc=\dfrac{[CO_2]}{[C][O_2]}[/tex]Further explanation

Given

Reaction

C+02 = CO2

Required

The equilibrium constant

Solution

The equilibrium constant is the ratio of concentration or pressure between the product and the reactant with each reaction coefficient raised  

The equilibrium constant is based on the concentration (Kc) in a reaction  

pA + qB -----> mC + nD  

[tex]\large {\boxed {\bold {Kc ~ = ~ \frac {[C] ^ m [D] ^ n} {[A] ^ p [B] ^ q}}}}[/tex]

So for the reaction :

C+O₂ ⇔ CO₂

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Answers

Answer:

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[tex]\Large \boxed{\sf by \ their \ properties}[/tex]

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A. What was the theoretical vield of C?
B. How much B was consumed by the reaction?

Answers

Answer:

A. Theoretical yield of C is 9.03 g

B. Mass of B consumed is 5.53 g

Explanation:

A. Determination of the theoretical yield of C.

Actual yield of C = 6.5 g

Percentage yield of C = 72.%

Theoretical yield of C =?

Percentage yield = Actual yield /Theoretical yield × 100

72% = 6.5 / Theoretical yield

72 / 100 = 6.5 / Theoretical yield

Cross multiply

72 × Theoretical yield = 100 × 6.5

72 × Theoretical yield = 650

Divide both side by 72

Theoretical yield = 650 / 72

Theoretical yield = 9.03 g

Therefore, the theoretical yield of C is 9.03 g

B. Determination of the mass of B consumed.

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

A + B —> C

Mass of A = 3.5 g

Mass of C = 9.03 g

Mass of B =?

A + B = C

3.5 + B = 9.03

Collect like terms

B = 9.03 – 3.5

B = 5.53 g

Thus, the mass of B consumed in the reaction is 5.53 g

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Answers

The volume required : 0.2 L

Further explanation

Given

500ml of 0.1M potassium dichromate (vii)

0.25 solution

Required

The volume

Solution

We can use dilution formula :

M₁V₁=M₂V₂

M₁ = 0.25

M₂ = 0.1

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V₁=(M₂V₂)/M₁

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V₁=0.2 L

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Answers

Answer:

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Answers

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Answers

Answer:

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OA. The atoms in a substance start to move faster until they are no longer touching each other.
OB.
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Answers

The option in “the mass of 1me of carbon equals the mass of 1 mole of boron atoms” is false since

Mass=molesx molar mass

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The number of atoms in 1 mole of carbon equals the number of atoms in 1 mole of boron is false because the molar mass of carbon is different from that of boron. Option A is correct.

The number of moles of an element is expressed as the ratio of the mass of the substance to its molar mass.

Moles  = Mass/Molar mass

This shows that the number of moles of the substance is dependent on the molar mass of the substance.

From the listed option the number of atoms in 1 mole of carbon equals the number of atoms in 1 mole of boron is false because the molar mass of carbon is different from that of boron

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