A solution is made containing 50.0 g of ethanol (C2H5OH) in 1000 g H2O. Calculate (a) the mole fraction of C2H5OH, (b) the mass percent of C2H5OH, (c) the molality of C2H5OH.
1. A solution is made containing 50.0 g of ethanol (C2H5OH) in 1000 g H2O. Calculate (a) the mole fraction of C2H5OH, (b) the mass percent of C2H5OH, (c) the molality of C2H5OH.
(a) SOLUTION:
Step 1: List the given values.
[tex]\begin{aligned} & mass_{\text{solute}} = \text{50.0 g} \\ & mass_{\text{solvent}} = \text{1000 g} \\ & MM_{\text{solute}} = \text{46.068 g/mol} \\ & MM_{\text{solvent}} = \text{18.015 g/mol} \end{aligned}[/tex]
Step 2: Calculate the number of moles of each component.
• For the solute (ethanol)
[tex]\begin{aligned} n_{\text{solute}} & = \frac{mass_{\text{solute}}}{MM_{\text{solute}}} \\ & = \frac{\text{50.0 g}}{\text{46.068 g/mol}} \\ & = \text{1.08535 mol} \end{aligned}[/tex]
• For the solvent (water)
[tex]\begin{aligned} n_{\text{solvent}} & = \frac{mass_{\text{solvent}}}{MM_{\text{solvent}}} \\ & = \frac{\text{1000 g}}{\text{18.015 g/mol}} \\ & = \text{55.5093 mol} \end{aligned}[/tex]
Step 3: Calculate the mole fraction of the solute.
[tex]\begin{aligned} X_{\text{solute}} & = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} \\ & = \frac{\text{1.08535 mol}}{\text{1.08535 mol + 55.5093 mol}} \\ & = \boxed{0.01918} \end{aligned}[/tex]
Hence, the mole fraction of ethanol is 0.01918.
------------------------------------------------------------
(b) SOLUTION:Step 1: List the given values.
[tex]\begin{aligned} & mass_{\text{solute}} = \text{50.0 g} \\ & mass_{\text{solvent}} = \text{1000 g} \end{aligned}[/tex]
Step 2: Calculate the mass percent of the solute (ethanol).
[tex]\begin{aligned} \%(m/m) & = \frac{mass_{\text{solute}}}{mass_{\text{solute}} + mass_{\text{solvent}}} \times 100\% \\ & = \frac{\text{50.0 g}}{\text{50.0 g + 1000 g}} \times 100\% \\ & = \boxed{4.76\%} \end{aligned}[/tex]
Hence, the mass percent of ethanol is 4.76%.
------------------------------------------------------------
(c) SOLUTION:Step 1: List the given values.
[tex]\begin{aligned} & mass_{\text{solute}} = \text{50.0 g} \\ & mass_{\text{solvent}} = \text{1000 g = 1.00 kg} \\ & MM_{\text{solute}} = \text{46.068 g/mol} \\ \end{aligned}[/tex]
Step 2: Calculate the number of moles of solute (ethanol).
[tex]\begin{aligned} n_{\text{solute}} & = \frac{mass_{\text{solute}}}{MM_{\text{solute}}} \\ & = \frac{\text{50.0 g}}{\text{46.068 g/mol}} \\ & = \text{1.08535 mol} \end{aligned}[/tex]
Step 3: Calculate the molality.
[tex]\begin{aligned} m & = \frac{n_{\text{solute}}}{mass_{\text{solvent}}} \\ & = \frac{\text{1.08535 mol}}{\text{1.00 kg}} \\ & = \text{1.08535 mol/kg} \\ & \approx \boxed{1.085 \: m} \end{aligned}[/tex]
Hence, the molality of ethanol is 1.085 m.
[tex]\\[/tex]
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2. In glucose fermentation, C2H5OH and CO2 are produced: C6H12O6 - 2 C2H5OH + 2CO2How many grams of CO2 are formed when 2.82 g of C2H5OH.
Answer:
2.69 g
Explanation:
Molecular weights
C = 12.01 g/mol
H = 1.008 g/mol
O = 16.00 g/mol
C2H5OH = 2(12.01) + 6(1.008) + 16.00 = 46.07 g/mol
CO2 = 12.01 + 2(16.00) = 44.01 g/mol
Starting with 2.82 g of C2H5OH
mass CO2 = 2.82 g C2H5OH × (1 mol C2H5OH / 46.07 g C2H5OH) × (2 mol CO2 / 2 mol C2H5OH) × (44.01 g CO2 / 1 mol CO2)
mass CO2 = 2.69 g
3. Calculate the mole fraction of c2h5oh in a solution that contains 46 grams of ethanol, c2h5oh, and 64 grams of methanol, ch3oh
SOLUTION:
Step 1: List the given values.
[tex]\begin{aligned} & mass_{\text{ethanol}} = \text{46 g} \\ & MM_{\text{ethanol}} = \text{46.069 g/mol} \\ & mass_{\text{methanol}} = \text{64 g} \\ & MM_{\text{methanol}} = \text{32.042 g/mol} \end{aligned}[/tex]
Step 2: Calculate the number of moles of ethanol and methanol.
• For ethanol
[tex]\begin{aligned} n_{\text{ethanol}} & = \frac{mass_{\text{ethanol}}}{MM_{\text{ethanol}}} \\ & = \frac{\text{46 g}}{\text{46.069 g/mol}} \\ & = \text{0.9985 mol} \end{aligned}[/tex]
• For methanol
[tex]\begin{aligned} n_{\text{methanol}} & = \frac{mass_{\text{methanol}}}{MM_{\text{methanol}}} \\ & = \frac{\text{64 g}}{\text{32.042 g/mol}} \\ & = \text{1.9974 mol} \end{aligned}[/tex]
Step 3: Calculate the mole fraction of ethanol
[tex]\begin{aligned} X_{\text{ethanol}} & = \frac{n_{\text{ethanol}}}{n_{\text{methanol}} + n_{\text{ethanol}}} \\ & = \frac{\text{0.9985 mol}}{\text{1.9974 mol + 0.9985 mol}} \\ & = \boxed{0.333} \end{aligned}[/tex]
Hence, the mole fraction of ethanol is 0.333.
[tex]\\[/tex]
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4. antibacterial effects of c2h5oh of carica papaya leaves
Answer:
The antibacterial activity of water and ethanol extracts of male and female Carica papaya leaves was carried out on pathogenic isolates. ... The ethanol extracts of both male and female papaya leaves demonstrated higher activities against all the Gram negative bacteria than Gram positive bacteria tested.5. Calculate the molar mass of c2h5oh if 8.0 mol of c6h12o6
Answer:
Molar mass of ethanol (c2h5oh)
=2(12)+6(1)+16.1=46.1g/mol
Explanation:
follow please
6. The complete combustion of ethanol, c2h5oh(l), to form h2o(g) and co2(g) at constant pressure releases 1235 kj of heat per mole of c2h5oh.
Answer: sorry po kailangan lang po ng points sorry po
Explanation: sorry po kailangan lang po ng points sorry po
7. what is the common name of C2H5OH
Hello? Can you give m the pictire?
Ethanol is the common name8. What is the molal concentration of 30 % ethanol solution (c2h5oh)?
Answer:
The molality of the solution is 8.919 molal.
Explanation:
Molality (m) is defined as moles of solute per kilogram of solvent. On the other hand, molarity (M) is a unit of concentration which has a unit of moles of solute per liters of solution.
Let us keep these in mind as we go further to the solution for this problem.
Given:
5.86 M (molar) of ethanol (C₂H₅OH)
Required:
Molality (m)
Solution:
We assume 1 liters of solution for this problem.
Get first the molar mass (MM) of the ethanol.
MM ethanol = (12.012 g/mol*2) + (1.008*5) + (15.999) + (1.008)
MM ethanol = 46.071 g/mol
Getting the mass of ethanol from 5.86 M, we have,
5.86 moles ethanol (since we assume 1 L of solution)
5.86 moles ethanol * 46.071 g/mol = 269.9761 grams ethanol
Getting the mass of solution from the density, we have,
0.927 g/mL * 1000 mL = 927 grams solution
(We used multiplier 1000 mL because 1 L = 1000 mL)
Solving for the mass of solvent, we have,
mass of solvent = mass of solution - mass of solute
mass of solvent = 927 grams - 269.9761 grams
mass of solvent = 657.024 grams solvent = 0.657024 kilograms solvent
Solving for molality, we have,
m = moles solute ÷ kilograms of solvent
m = 5.86 moles of ethanol ÷ 0.657024 kg solvent
m = 8.919
Therefore, the molality of the solution is 8.919 molal.
Answer:
7.34 m
Explanation:
Answer can be found in this link: https://www.assignmentexpert.com/homework-answers/chemistry/organic-chemistry/question-137007
Paki brainliest thank you!
9. A 21.8 g sample of ethanol (c2h5oh) is burned
Answer:
The specific heat capacity of combustion of 21.8 of ethane is \rmm \bold { 718. 394 J/kg^ \cdot C}\rmm718.394J/kg
⋅
C
Given here,
The molar mass of ethanol is 46.07 g/mol.
Enthalpy of the combustion \rm \bold{ \Delta H = -1235 kJ}ΔH=−1235kJ
Hence the number of moles in 21.8 g of ethanol is 0.473 mole
The energy released during reaction
\rm \bold{ Q = n \times \Delta H}Q=n×ΔH
Q = - 584.15 kJ
As we know Heat released by ethanol is equal to hat absorbed by calorimeter.
The heat absorbed = 584.15 kJ
The specific heat capacity formula
\rm \bold{ c= \frac{Q}{m \times \Delta T} }c=
m×ΔT
Q
Where,
Q = heat absorbed = 584155 J
m = mass = 21.8g
\rm \bold{ \Delta T} } - change in temperature = 37.3 °C
\rm \bold{ c= \frac{584.15 kJ}{ 21.8g \times 37.3} }c=
21.8g×37.3
584.15kJ
\rmm \bold { c = 718. 394 J/kg^ \cdot C}\rmmc=718.394J/kg
⋅
C
10. BALANCE CHEMICAL EQUATION OF C2H5OH + O2 > CO2 + H2O
Answer:
C2H5OH + 3O2 > 2CO2 + 3H2O
Explanation:
In the Picture
11. what is the mollar masses ofH20NaCIC2H5OHCO2mg
Answer:
18.01528 g/mol
58.44 g/mol
46.07 g/mol
44.01 g/mol
24. 305 g/mol.
Answer:
molecular mass of H₂= 2× atomic mass of H
= 2×1 = 2 u.
molecular mass of O₂= 2× atomic mass of O
= 2 × 16u = 32 u.
molecular mass of Cl₂= 2 × atomic mass of Cl
= 2 × 35.5 u= 71 u.
molecular mass of CO₂= atomic mass of C + 2 × atomic mass of O
= 12 + 2×16
= 12+ 32= 44 u.
molecular mass of CH₄= atomic mass of C + 4 × atomic mass of H
= 12+ 4×1= 16 u.
molecular mass of C₂H₆= 2 × 12 + 6 × 1
= 30 u.
molecular mass of C₂H₄= 2 × 12 + 4 × 1
= 28 u.
molecular mass of NH₃= 14+ 3×1
= 17 u.
molecular mass of CH₃OH= 12 + 3× 1+ 16 × 1
= 32 u.
#Lets learncorrect me if i a wrong12. 2. The boiling point of Ethanol (C2H5OH) is 78°C, while the Sodium Chloride (NaCl)has 1420°C. Which of the following is the correct statement in determining thesecompounds?A. C2H5OH - covalent; NaCl-ionic C. both are ionicB. C2H5OH - ionic; NaCl- covalent D. both are covalent
Answer:
A. C2H5OH - covalent; NaCl - ionic
basta yan yung sagot ko
13. Calculate the molal concentration of 40% ethanol solution (c2h5oh).
SOLUTION:
Step 1: List the given values.
[tex]\begin{aligned} & MM_{\text{solute}} = \text{46.069 g/mol} \\ & \%(m/m) = 40\% \end{aligned}[/tex]
Step 2: Calculate the mass of solute (ethanol).
Assume that the mass of solution is 100 g.
[tex]\begin{aligned} mass_{\text{solute}} & = \frac{\%(m/m)}{100\%} \times mass_{\text{solution}} \\ & = \frac{40\%}{100\%} \times \text{100 g} \\ & = \text{40 g} \end{aligned}[/tex]
Step 3: Calculate the number of moles of solute.
[tex]\begin{aligned} n_{\text{solute}} & = \frac{mass_{\text{solute}}}{MM_{\text{solute}}} \\ & = \frac{\text{40 g}}{\text{46.069 g/mol}} \\ & = \text{0.86826 mol} \end{aligned}[/tex]
Step 4: Calculate the mass of solvent.
[tex]\begin{aligned} mass_{\text{solvent}} & = mass_{\text{solution}} - mass_{\text{solute}} \\ & = \text{100 g} - \text{40 g} \\ & = \text{60 g} \\ & = \text{0.0600 kg} \end{aligned}[/tex]
Step 5: Calculate the molality of solution.
[tex]\begin{aligned} m & = \frac{n_{\text{solute}}}{mass_{\text{solvent}}} \\ & = \frac{\text{0.86826 mol}}{\text{0.0600 kg}} \\ & = \text{14.5 mol/kg} \\ & = \boxed{14.5 \: m} \end{aligned}[/tex]
Hence, the molal concentration of 40% ethanol solution is 14.5 m.
[tex]\\[/tex]
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14. 2. The boiling point of Ethanol (C2H5OH) is 78°C, while the Sodium Chloride (NaCl)has 1420°C. Which of the following is the correct statement in determining thesecompounds?A. C2H5OH - covalent; NaCl- ionic C. both are ionicB. C2H5OH - ionic; NaCl- covalent D. both are covalent
Answer:
My answer is
A. C2H5OH - covalent; NaCI- ionic
15. C2H5OH what is the molar mass
Answer:
C2H5OH/ Ethanol
Molar mass: 46.07 g/mol
Explanation:
16. calculate molar mass Ethyl alcohol, C2H5OH
Answer:
46.07 G/MolExplanation:
Basta Tama yan
17. antibacterial effects of c2h5oh
Screening of natural extracts is a focused intensive study that aims to find active principles sorted from plant resources both safe and environmental friendly. The present study was aimed to evaluate the antibacterial activity of direct crude extracts of Carica papaya leaves and Allium sativum cloves alone and in combination against multiple drug resistant strains
HOPE IT HELP:-)
18. how many atoms are there in ethyl alcohol C2H5OH
Answer:
two carbon atoms six hydrogen atoms and one oxygen atom.
19. 3. What is the molarity of a solution containing 115g of ethanol (C2H5OH)?
Answer:
The density of an aqueous solution containing 10% of ethanol, C2H5OH, by mass is 0.984 g/mL. Calculate the a) M b) m of this equation c) What volume of the solution would contain 0.125 mole of ethanol? d) Calculate the mole fraction of water inthesolution
20. What is the total number of moles contained in 115 grams of C2H5OH?
The total number of moles : 2.5
Further explanationAtoms are composed of 3 types of basic particles namely protons, electrons, and neutrons
the mass of this basic particle is expressed in atomic mass units (amu)
This atomic unit uses the standard atomic mass, that is, the C-12 isotope
1 atom C-12 = 12 atomic mass units
1 atomic mass unit = 1/12 x mass 1 C-12 atom
Mole itself is the number of particles contained in a substance
1 mole = 6.02.10²³ particles
Mole can be determined if the amount of substance mass and its molar mass is known
[tex]\tt \boxed{\bold{n=\dfrac{m}{M}}}}[/tex]
Given
mass = 115 grams
Required
The number of moles
Solution
the molar mass of C₂H₅OH :
= 2.Ar C+ 6.Ar H + Ar O
= 2.12 + 6.1 + 16
= 46 g/mol
moles :
[tex]\tt =\dfrac{115}{46}=\boxed{\bold{2.5}}[/tex]
Learn morethe mole fractions of the solute and the solvent
https://brainly.ph/question/2062141
moles of CaCO3
https://brainly.ph/question/2356002
the mole ratio of Al to Cl2
https://brainly.ph/question/2124499
#LetsStudy
21. Ethanol, C2H5OH, is found in gasoline blends used in many parts of North America. Write a balanced chemical equation for the combustion of C2H5OH to form CO2 and H2O.
Answer:
C2H5OH + 3O2. 2CO2 + 3H2O
Explanation:
Ethanol react with oxygen to produce carbon dioxide and water. Ethanol combustion on air
22. How many moles are represented by 11.5 g of C2H5OH
Hello!
How many moles are represented by 11.5 g of C2H5OH ?
C = 2*(12u) = 24 u
H = 6*(1u) = 6 u
O = 1*(16u) = 16 u
---------------------------
molecular mass of C2H5OH (or C2H6O) = 24 + 6 + 16 = 46 g/mol
How many moles are represented by 11.5 g of C2H5OH ?
46 g ---------------- 1 mol
11.5 g ------------- y mol
Product of extremes equals product of means
[tex]46*y = 11.5*1[/tex]
[tex]46\:y = 11.5[/tex]
[tex]y = \dfrac{11.5}{46}[/tex]
[tex]\boxed{\boxed{y = 0.25\:moles\:of\:C_2H_5OH}}\end{array}}\qquad\checkmark[/tex]
________________________
I Hope this helps, greetings ... Dexteright02! =)
23. A 21.8 g sample of ethanol (c2h5oh) is burned
Answer:
it will make a fire
Explanation:
since ethanol has condiments same as alcohol
24. what is the moral mass of c2h5oh
Answer:
46.07 g/mol
Explanation:
Formula: C2H5OH
IUPAC ID: ethanol
Density: 789 kg/m³
25. 2. The boiling point of Ethanol (C2H5OH) is 78°C, while the Sodium Chloride (NaCl) has 1420°C. Which of the following is the correct statement in determining these compounds? A. C2H5OH - covalent; NaCl- ionic C. both are ionic B. C2H5OH - ionic; NaCl- covalent D. both are covalent
Answer:
bored ako door open
Explanation:
i dont know
26. How many moles are in 23 grams of ethanol, C2H5OH?
Explanation:
23 g ethanol is 23/46 =0.50 mole ethanol Burning 1.0 mole ethanol gives 2 mole of carbon dioxide, because there are two C in one C2H5OH So 0.50 mole ethanol produces 1.0 mole carbon dioxide, that is 44 gram.
27. how many grams of CO2 will be produced from 18.0g of C2H5OH
792 grams of CO2 will be produced
28. Why ethanol (c2h5oh) is not soluble in cyclohexane (c6h12)?
C2H5OH has strong Hydrogen bond.
29. how many atoms are there in the hydrogen (H) of ethanol (C2H5OH)
Answer:
A Sample Of Ethanol (ethyl Alcohol), C2H5OH, Contains 4.2 X 10^23 Hydrogen Atoms. Find answers now! In on molecule of C2H5OH there are 9 atoms.
Explanation:
30. common name of C2H5OH
Answer:
ETHANOLEthanol, a volatile, flammable, colorless liquid with the molecular formula of C2H5OH, is most commonly known as alcohol.
hope it helps.
Answer:
ETHANOL
Explanation:
Ayun lang po aalam ko pakitama nalang po kung mali
