18 mL of solvent was added to 82 mL of distilled water for a final solution volume of 100 ml. What is the percent concentration of solvent in the final solution?

Answer:

Explanation:

If our entire Solar System were the size of a quarter, the planets and the sun is now a tiny speck of dust. The flat disc of the coin can represent the orbits of the planets.

Using this scale, the diameter of our Milky Way galaxy will be about the size of North America.

The nearest star, other than our own Sun, is about four light years away. That means it takes four years for its light to reach us. Since light travels at a speed of 3.0 x 10^8 meters per second, each light year is such a great distance. Proxima Centauri Milky way would be another quarter, two soccer fields away.

A much further star, Deneb is actually 1,800 light years away, the nearest star would be about 24,000 miles away.

Answer:

The answer to this question is 45.33 moles of aluminum sulfate is produced when 125 moles of aluminum hydroxide and 136 moles of sulfuric acid react

Explanation:

To solve this, we write out the chemical equation of he reaction thus

Al(OH)3(s) + 3 H2SO4(aq) -----> Al2 (SO4)3(aq) + 6 H2O(l)

here it is seen that one moles of aluminum hydroxide reacts with three moles of sulfuric to produce one mole of aluminum sulfate and six moles of water

hence

136 moles of sulfuric acid reacts with 136/3 or 45.33 moles of aluminum hydroxide to produce 136/3 or 45.33 moles of aluminum sulfate and 2× 136 moles of water

Hence the amount in moles of aluminum sulfate produced is 45.33 moles

Final answer:

Using the balanced chemical equation 3 Al(OH)3 + 3 H2SO4 → Al2(SO4)3 + 6 H2O, and knowing that aluminum hydroxide is the limiting reactant, 125 moles of aluminum hydroxide will produce 41.67 moles of aluminum sulfate.

Explanation:

The question asks how many moles of aluminum sulfate will be produced when reacting 125 moles of aluminum hydroxide with 136 moles of sulfuric acid. To answer this, we need the balanced chemical equation:

3 Al(OH)3 + 3 H2SO4 → Al2(SO4)3 + 6 H2O

The stoichiometry of the reaction shows that 3 moles of aluminum hydroxide react with 3 moles of sulfuric acid to produce 1 mole of aluminum sulfate. Since there are more moles of sulfuric acid present, aluminum hydroxide is the limiting reactant. Therefore, we can calculate the moles of aluminum sulfate produced by dividing the moles of aluminum hydroxide by 3, which gives us:
125 moles Al(OH)3 ÷ 3 = 41.67 moles Al2(SO4)3 (rounded to two decimal places as per the significant figures in the provided moles of reactants).

Answer:

The answer to your question is below

Explanation:

The specific heat is a physical property equal to the amount of heat necessary to increase the temperature of 1 gram of a substance by one degree celsius.

The lower the specific heat, the lower the amount of heat to increase the temperature 1°C, the higher the specific heat, the higher the amount of heat necessary to increase the temperature by 1°C.

The specific heat of copper is 0.093 cal/g°C

The specific heat of water is 1 cal/g°C.

That is why is necessary more heat to warm water.

More energy to increase the temperature of 100 grams of liquid water by one degree Celsius than it does 100 grams of copper metal due to higher specific heat.

Specific heat refers to the amount of heat needed to raise the temperature of 1 gram of a substance by 1 degree Celsius. Water has a high specific heat which means it takes more energy to increase the temperature of water compared to other substances like metals.

So we can conclude that more energy is needed to increase the temperature of 100 grams of liquid water by one degree Celsius than copper metal because of higher specific heat of water.

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Answer:

To regulate the gas pressure in the lighter tank and avoid build-up of pressure.  

Explanation:

We soak the lighter in the water bath to achieve uniform heating and reduce the fire hazard inherent in direct heating methods, ensuring a safer laboratory environment.

We soak the lighter in the water bath to ensure uniform heating of the reaction mixture with less fire hazard. Direct heating on a Bunsen burner or hot plate can cause uneven temperatures and increased risks of fire or overheating. A water bath is especially beneficial when a chemical reaction must be heated for a certain time to occur. It provides a stable and consistent heat source, which is safer and can prevent accidents in the lab. By using a water bath, we also prevent the lighter from getting excessively hot, which could cause it to malfunction or pose a safety risk.

Phosphoric acid is a triprotic acid that can donate three protons in solution, forming three anions. The pH of a 0.100 M solution is approximately 1.0, assuming it only ionizes once. The concentrations of the formed species are estimated to be highest for H2PO4⁻ and much lower for HPO4²⁻ and PO4³⁻.

Phosphoric acid, H3PO4 is a triprotic acid, meaning it can donate three hydrogen ions in a solution. This results in the formation of three different species:  H2PO4⁻, HPO4²⁻, and PO4³⁻.

The estimated pH of a 0.100 M phosphoric acid solution will depend on the degree of dissociation, but for the first ionization, we can approximate it using the expression pH=-log[H⁺], where [H⁺] is the hydronium ion concentration. Given that a 0.100 M phosphoric acid solution ionizes mostly once, we have [H⁺]≈0.100 M, leading to an estimated pH around 1.0.

Next, the concentrations of the species in equilibrium can be calculated without exact Kb or Ka values as long as we make the approximation that dissociation after the first hydrogen ion is minimal in a dilute solution like 0.100 M. In this case, we will assume [H2PO4⁻]≈0.100 M and [HPO4²⁻] and [PO4³⁻] will be much less than [H2PO4⁻].

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Answer:

True

Explanation:

This are acts or actions that concur with situational expectations.

Answer:

The right answer to this question is no, all of the liquid will not evaporate, there will be 8.61 ×10⁻⁴ moles or 6.22×10⁻² grams left in the container

At equilibrium the pressure in the container will be the vapor pressure of liquid pentane which is = 100. mm Hg

Explanation:

To solve this we list the known values as follows

vapor pressure of liquid pentane = 100 mmHg = 13.33 KPa

Temperature T = 260 K

Volume of container = 350 mL = 0.00035 m³

The number of moles of liquid pentane = n

The universal gas constant = R = 8.314 J/(mol·K)

Thus From the ideal gas equation PV = nRT →

Thus plugging in the values in the above equateion we have

n = [tex]\frac{PV}{RT} = \frac{(13330)(0.00035)}{(8.314)(260)}[/tex] = 2.16×10⁻³ moles

Hence the number of moles in 0.218 g sample of liquid   pentane C₅H₁₂ with molar mass = 72.15 g/mol = 0.218/72.15 = 3.02×10⁻³ moles

Hence the number of moles present in the sample placed in the closed evacuated container = 3.02×10⁻³ moles

However number of moles to completely evaporate at 100 mmHg and 260 K is 2.16×10⁻³ moles hence, 3.02×10⁻³ moles - 2.16×10⁻³ moles,  or 8.61 ×10⁻⁴ moles will be left in the container

converting the value in moles to mass we have number of moles, n = mass/(molar mass)

Therefore the mass = number of moles × molar mass = 8.61 ×10⁻⁴ × 72.15 = 6.22 × 10⁻² grams left in the container

The pressure in the container at equilibrium will be vapor pressure of liquid pentane C₅H₁₂, or 100. mm Hg

In the described closed system, all of the liquid pentane will evaporate due to the vapor pressure. The final pressure when equilibrium is reached will be the vapor pressure, which is 100.0 mm Hg.

The subject of the question is the vapor pressure of pentane, C5H12, which is a concept from Chemistry.

The given vapor pressure is 100.0 mm Hg at 260 K. In this closed system the liquid and vapor will come to equilibrium at the given temperature, at which point the vapor pressure will be equal to the given 100.0 mm Hg.

The sample mass of 0.218 g doesn't exceed the amount needed for evaporation at the equilibrium pressure. Therefore, all of the liquid pentane will evaporate.

The final pressure in the container when equilibrium is reached will be the vapor pressure, which is 100.0 mm Hg.

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Answer : The pressure of the helium gas inside the container would be, 32.1 atm

Explanation :

To calculate the pressure of the gas we are using ideal gas equation as:

[tex]PV=nRT[/tex]

where,

P = Pressure of [tex]He[/tex] gas = ?

V = Volume of [tex]He[/tex] gas = 7.5 L

n = number of moles [tex]He[/tex] = 10 mole

R = Gas constant = [tex]0.0821L.atm/mol.K[/tex]

T = Temperature of [tex]He[/tex] gas = [tex]20^oC=273+20=293K[/tex]

Putting values in above equation, we get:

[tex]P\times 7.5L=10mole\times (0.0821L.atm/mol.K)\times 293K[/tex]

[tex]P=32.1atm[/tex]

Thus, the pressure of the helium gas inside the container would be, 32.1 atm

The net ionic equation for the reaction of strontium chloride and potassium sulfate, forming strontium sulfate solid, is Sr2+(aq) + SO42-(aq) → SrSO4(s)

When an aqueous solution of strontium chloride is mixed with an aqueous solution of potassium sulfate, strontium sulfate precipitates out and potassium and chloride ions remain in the solution. The balanced net ionic equation for this reaction is as follows:

Sr2+(aq) + SO42-(aq) → SrSO4(s)

This equation represents the change where strontium ions from strontium chloride and sulfate ions from potassium sulfate are combined to form strontium sulfate solid. The charges are balanced with the positive 2 charge of the strontium ion balancing the negative 2 charge of the sulfate ion.

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Answer:

                      ΔT  =  20.06 °C

Explanation:

The equation used for this problem is as follow,

                                    Q  =  m Cp ΔT   ----- (1)

Where;

           Q  =  Heat  =  1.17 kJ = 1170 J

           m  =  mass  =  24.1 g

           Cp  =  Specific Heat Capacity  =  2.42 J.g⁻¹.°C⁻¹

           ΔT  =  Change in Temperature  =  ??

Solving eq. 1 for ΔT,

                                ΔT  =  Q / m Cp

Putting values,

                                ΔT  =  1170 J / 24.1 g × 2.42 J.g⁻¹.°C⁻¹

                                ΔT  =  20.06 °C

Answer:

The answer to this is the concentration of glucose (C6H12O6) in mg/dL = 95.48 mg/dL

Explanation:

To solve this we list out the known variables thus

Measured concentration of  glucose  (C6H12O6) = 5.3mM per liter

The molar mass of glucose  = 180.156 g/mol

From the above, it is seen that one mole of glucose contains 180.156 grams of  C6H12O6 therefore 5.3 mM which is 5.3 × 10⁻³ moles contains

5.3 × 10⁻³ moles × 180.156 g/mol = 0.9548 grams of glucose

Also 1 d L = 0.1 L  or 1 L = 10 dL and 1 mg = 1000 g, hence

thus 0.9548 grams per liter is equivalent to 1000/10 × 0.9548 milligrams per dL or 95.48 mg/dL

Answer:

The complete question is:

Question: What disorder is indicated by these findings? A client comes to the emergency department with status asthmaticus. His respiratory rate is 48 breaths/minute, and he is wheezing. An arterial blood gas analysis reveals a pH of 7.52, a partial pressure of arterial carbon dioxide (PaCO2) of 30 mm Hg, PaO2 of 70 mm Hg, and bicarbonate (HCO3−) of 26 mEq/L.

A. Metabolic acidosis

B. Respiratory acidosis

C. Metabolic alkalosis

D. Respiratory alkalosis

Answer: The correct answer is:

D. Respiratory alkalosis

Explanation:

In Respiratory alkalosis the Partial Pressure of Arterial Carbondioxide (PaCO2) become decreased (i.e. less than 35 mm Hg) and the pH of blood become increased (i.e. more than 7.45). Alveolar hyperventilation causes respiratory alkalosis.

Alveolar hyperventilation occurs when alveolar ventilation is increased than the arterial carbondioxide tension and carbondioxide production.

Alveolar ventilation is the gaseous exchange between alveoli and the external environment.

Whereas, in metabolic acidosis, bicarbonate (HCO3) become decreased (i.e. less than 22 mEq/l and the pH of blood become decreased (i.e. less than 7.35); in respiratory acidosis,  the pH of blood also become decreased (i.e. less than 7.35) and the PaCO2 become increased (i.e. more than 45 mm Hg); and in metabolic alkalosis,  the bicarbonate (HCO3) become increased (i.e. more than 26 mEq/l and the pH become increased (i.e. more than 7.45).

Answer:

D (or E If properly listed to include the active site option)

Explanation:

A. Is true

Enzymes are organically biochemical catalyst and thus they can speed up the rate of chemical reaction in the body

B is true

They are catalysts as said earlier

C is true

They have active sites. An enzyme does not act on all substrates. They have particular group on which they can act. For example, we have carbohydrates enzymes that act on carbohydrates substrate only. This enzymes have no business acting on a protein substrate.

D. Enzymes are proteins

One of the important characteristics of enzymes is that they are protenious in nature

E. This is wrong. Enzymes like any over catalyst are not consumed in the course of the biochemical reaction

Enzymes are biological catalysts that speed up chemical reactions. These are proteins and have an 'active site' where reactions occur. However, they are not consumed during the reaction.

In context to your question about enzymes, they serve specific roles as biological accelerators or catalysts, significantly boosting the rate of a chemical reaction. This property is showcased under option A. They are indeed classified as proteins (option C), and they do have parts called active sites, where the substrate (the molecule upon which the enzyme acts) binds (option B).

However, option D posits that enzymes are consumed during the reaction, which is incorrect. Unlike many catalysts in non-biological reactions that get consumed during the reaction, enzymes remain unaffected by the reaction. They don't exhaust or alter in a reaction and are available to facilitate other reactions once the process is finished.

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Answer:

Percent composition of the solution is 26 % of sucrose and 74 % of water

Explanation:

Percent composition is the mass of solute, either of solvent in 100 g of solution.

Mass of solution = Mass of solvent + Mass of solute

Mass of solute = 35 g

Mass of solvent = 100 g

As we know, water density = 1g/mL

So 1g/mL . 100 mL = 100 g

35 g + 100 g = 135 g → Mass of solution

(Mass of solute / Mass of solution) . 100 =

(35 g / 135 g) . 100 = 26 %

(Mass of solvent / Mass of solution) . 100 =

(100 g / 135 g) . 100 = 74 %

To calculate the percent composition of sucrose in the solution, divide the mass of sucrose (35 grams) by the total mass of the solution (sucrose plus water, which is 135 grams) and multiply by 100%. The solution has a percent composition of approximately 25.93% sucrose.

The question involves calculating the percent composition of a solution by mass. If a solution contains 35 grams of sucrose (C12H22O11) in 100 mL of water (noting that the density of water is roughly 1 g/mL, so we have 100 grams of water), the total mass of the solution is the sum of the mass of the solute (sucrose) and the solvent (water), which is 35 g + 100 g = 135 g. To find the percent by mass of sucrose in the solution, we use the formula:

Percent by mass of sucrose = (Mass of sucrose / Total mass of solution) × 100%

Inserting the values we have:

Percent by mass of sucrose = (35 g / 135 g) × 100% ≈ 25.93%

Therefore, the percent composition of sucrose in the solution is approximately 25.93%.

Answer:

we only see parts of the lit side as the moon goes around the earth

Explanation:

Unlike the sun, the moon orbits the Earth. This is the reason why we see the different phases of the moon. The reflection of the moon is being illuminated back to us with the help of the sun. So, as the moon circles the Earth, we only see parts of the lit side. Such changes helps us see the moon in different phases such as the Third Quarter, Crescent, New Moon, Full Moon, etc.

For example, during "Full Moon," the moon's entire face is lit up by the sun. Thus, we see the entire moon's lit portion.

Thus, this explains the answer.

Answer: The average potential energy of the PIB is 0 irrespective of the wave function.

Explanation:

⟨H⟩=⟨KE⟩+⟨V⟩

the nn quantum number

⟨KE⟩=(π^2 ℏ^2)/(2mL^2 )

the average kinetic energy of the wavefunction is dependent on

⟨V⟩=∫sin(kx)0sin(kx)dx=0

The average potential energy of the PIB is 0 irrespective of the wave function.

⟨H⟩=⟨KE⟩=(π^2 ℏ^2)/(2mL^2 )

No I don't think it works that way. What you could do is drink one energy drink, wait 5 hours and then drink another.

Answer:

10 hrs of energy

Explanation:

these bottles hold energy for 5hrs so if it is double you have double the time of energy

Answer:

5.6 × 10⁻¹⁷ J

Explanation:

We know that the work done by an electric field, E on an electric charge, q moving a distance, d is W = qEdcosθ. From above, q = 3.2 × 10⁻¹⁹ C, d = 20 cm = 2 × 10⁻² m, E = V/d = 175V /2 × 10⁻² m = 8750 V/m and θ = 0 since the α particle moves in the same direction as the electric field. So W = qEdcosθ = 3.2 × 10⁻¹⁹ C × 8750 V/m × 2 × 10⁻² m × cos0  = 5.6 × 10⁻¹⁷ J. We know that the work done by the electric field on the charge W = ΔK the change in kinetic energy of the charge. So, W = 1/2m(v₂² - v₁²) where v₁ = initial velocity and v₂ = final velocity. Since the charge is at rest at the positive plate, v₁ = 0. So, W = 1/2mv₂² = K which is the kinetic energy of the particle after moving the distance of 20 cm between the plates. So K = W = 5.6 × 10⁻¹⁷ J

Answer:

the correct answer is d

Explanation:

The HA bar on the right must be converted completely to H3O+ and A-. Strong acids completely dissociate in solution. Complete dissociation would mean that there is no HA bar left on the right of the arrow.

To represent a strong acid in the graph, the 'HA' bar has to be almost non-existent while the H3O+ and A- bars significantly increase reflecting the fact that strong acids completely disassociate in water. The correct answer is 'The HA bar on the right must be converted completely to H3O+ and A-'.

The correct answer to the given question is option D).

In the context of this question, the graph represents a moderately weak acid and we're asked to determine what changes would occur in the graph for a relatively strong acid.

Here, the 'HA' would represent the weak acid that partially disassociates into H3O+ (hydronium ions) and A- (the conjugate base). One characteristic of a strong acid is that it completely disassociates in water.

Therefore, to represent a strong acid, the 'HA' bar on the right would need to be much lower or even non-existent (to indicate complete disassociation). In turn, the H3O+ and A- bars on the right would need to increase significantly/acquire all the 'HA' bar's original height to represent the products of the strong acid's complete disassociation.

So, the correct answer would be (D) 'The HA bar on the right must be converted completely to H3O+ and A-'.

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Answer:

The final temperature of the mixture is 21.4°C

Explanation:

Specific heat capacity of water = 1 cal/g°c

Heat loss by water  = 77.1 g X 1 cal/g°c X 50.9°c = 3924.39 Cal

Latent heat of fusion of ice = 79.7 g⁻¹

Heat required to melt ice at 0°c= 22.5 g X 79.7 g⁻¹ = 1793.25 Cal

Heat gained by ice from water at a higher temperature, T°c = 22.5 X 1 X T  

= 22.5T

Also heat lost by water = 77.1  X 1  X (50.9-T)cal

By calorimetric principle

Heat lost by a hot body = heat gained by a cold body

77.1  X 1  X (50.9-T) = 22.5T + 1793.25

3924.39 -77.1T = 22.5T + 1793.25

99.6T = 3924.39 - 1793.25

T = 2131.14/99.6

T = 21.4°C

Therefore, the final temperature of the mixture is 21.4°C

Answer: A. Perform a gravity filtration with a stemless funnel.

Explanation: Gravity filtration with a stemless funnel prevent blocking up the funnel. Undissolved sample may come out in the stem when the solution cools down thus will be blocking the funnel. Using stemless funnel prevent this problem.

Answer:

The first question is water vapor.

Explanation:

Water vapor is not a green house gas. Clouds are made of water vapor.

Answer:

65.4 is the mass for 1.9×10²⁴ atomsof Pb

Explanation:

1mol of atoms of Pb has → NA (6.02×10²³ atoms) and weighs → 207.2 g

Therefore 1.9×10²⁴ atomsof Pb may weigh (1.9×10²⁴ . 207.2) / NA = 65.4 g

Answer: The proteins were no longer soluble in the blood.

Answer:

Synergism

Explanation:

This is an example of Synergism. Synergism is nothing but working out of two medicines together.

Examples of medical synergies are when doctors treat microbial heart infections with ampicillin and Gentamicin and when people with cancer undergo radiation and chemotherapy or more than one chemotherapy drug at a time.

Answer: (a) NH3 + H20 ⇔ NH4⁺ + OH⁻ Kb =  [tex]\frac{ [NH4][OH]}{[NH3]}[/tex]

(b) CN⁻ + H20 ⇔ HCN⁺ + OH⁻ Kb =  [tex]\frac{ [HCN][OH]}{[CN]}[/tex]

(c) C5H5N + H20 ⇔ C5H5NH⁺ + OH⁻ Kb = [tex]\frac{[C5H5NH][OH]}{ [C5H5N]}[/tex]

(d) C6H5NH2 + H20 ⇔ C6H5NH3⁺ + OH⁻ Kb = [tex]\frac{[C6H5NH3][OH]}{[C6H5NH2] }[/tex]

Explanation: There are ways to calculate the strength of acids and bases. The pH is more commom but there is pKa, pKb, Ka and Kb. Ka and Kb are the dissociation constant for acids and bases, respectively.

Using the balanced equation of dissociation, Kb is calculated by dividing the concentration of the products with the concentration of the reagent.

The reactions and Kb equilibrium expressions for NH₃, CN⁻, pyridine (C₅H₅N), and aniline (C₆H₅NH₂) as bases in water are summarized including all necessary chemical equations and equilibrium expressions.

The following reactions and corresponding Kb equilibrium expressions for NH₃, CN⁻, pyridine (C₅H₅N), and aniline (C₆H₅NH₂) acting as bases in water are as follows:

Answer:

molecules

Explanation:

Answer:

Explanation:

The same charges repel each other while opposite charges attract each other. During electron-electron interaction repulsion take palace because the electron has negative charges. Nucleus has positive charges so the interaction between two nucleus results in the form of repulsion. When interaction takes place between nucleus and electron then attraction takes place between nucleus and electrons due to opposite charges.

          The formation of a bond that takes place due to the sharing of the electrons is known as a covalent bond and thus the covalent molecule is formed.

The forces between two atoms is the result of electron-electron repulsion, nucleus-nucleus repulsion, and nucleus-electron attraction. At the point of equilibrium, the attractive forces balance the repulsive forces. The most stable configuration of atoms exists at the point of minimum potential energy, when the atoms bond covalently and a covalent bond forms.

The force of attraction between atoms leads to the formation of covalent bonds. The force of attraction is defined by the magnitude of oppositely charged ions bonded with one another. There are various forces of attraction in between molecules which are studied in chemistry. These are:

Answer:

the velocity of the tomato will be  v  = u + gt

kinetic energy = 9.64 h

Explanation:

At that time, let the initial speed be u.

The acceleration due to gravity be g

The final velocity will be given as v = u + at

but a = g = 9.81 m/s² [acceleration due to gravity]

so the final velocity will be given as v = u + at

Let the potential energy be Ep

Before landing the ground, 93.8 % of the potential energy will be converted to kinetic energy. Therefore, the calculation will be as follows:

Ep = mgh

Kinetic energy Ek = 1/2mv²

But, Ek = 0.938 Ep

            = 0.983 × gh

            = 9.64 h

where h is the height of the object from the ground.

To calculate the vapor pressure of the solution, we need to consider Raoult's law. According to Raoult's law, the vapor pressure of a solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent. In this case, the solvent is water and the solute is K3PO4. To find the mole fraction of water, we divide the moles of water by the total moles of solute and solvent. Using the given values, the vapor pressure of the solution at 55 °C is 101.0 torr.

To calculate the vapor pressure of the solution, we need to consider Raoult's law. According to Raoult's law, the vapor pressure of a solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent. In this case, the solvent is water and the solute is K3PO4. To find the mole fraction of water, we divide the moles of water by the total moles of solute and solvent. Using the given values, we have:

Moles of water = 0.941 mol

Total moles of solute and solvent = 0.159 mol K3PO4 + 0.941 mol H2O = 1.1 mol

Mole fraction of water = (0.941 mol) / (1.1 mol) = 0.855

Now, we can use the mole fraction of water and the vapor pressure of pure water to calculate the vapor pressure of the solution:

Vapor pressure of the solution = mole fraction of water * vapor pressure of pure water

= 0.855 * 118.1 torr

= 101.0 torr

Therefore, the vapor pressure of the solution at 55 °C is 101.0 torr.

Using Raoult's Law, the vapor pressure of the solution at 55°C is approximately 100.93 torr.

To calculate the vapor pressure of the solution, we will use Raoult's Law, which states:

[tex]\[ P_{\text{solution}} = X_{\text{solvent}} \cdot P^{\star}_{\text{solvent}} \][/tex]

where:

- [tex]\( P_{\text{solution}} \)[/tex] is the vapor pressure of the solution,

- [tex]\( X_{\text{solvent}} \)[/tex] is the mole fraction of the solvent,

- [tex]\( P^{\star}_{\text{solvent}} \)[/tex] is the vapor pressure of the pure solvent.

Given data:

- Mole fraction of water [tex](\( X_{\text{H2O}} \))[/tex] in the solution:

 [tex]\[ X_{\text{H2O}} = \frac{n_{\text{H2O}}}{n_{\text{H2O}} + n_{\text{K3PO4}}}[/tex]

[tex]= \frac{0.941}{0.159 + 0.941} \\\\= \frac{0.941}{1.1} \\ \\ \approx 0.855[/tex]

- Vapor pressure of pure water at 55 °C:

[tex]\[ P^{\star}_{\text{H2O}} = 118.1 \text{ torr} \][/tex]

Now, calculate the vapor pressure of the solution:

[tex]\[ P_{\text{solution}} = X_{\text{H2O}} \cdot P^{\star}_{\text{H2O}} \][/tex]

[tex]\[ P_{\text{solution}} = 0.855 \times 118.1 \text{ torr} \][/tex]

[tex]\[ P_{\text{solution}} \approx 100.93 \text{ torr} \][/tex]

Therefore, the vapor pressure of the solution at 55 °C is approximately 100.93 torr.