19.11.2021

What is the molecular mass for a non-electrolyte if 35.0 g of it is dissolved in 45.0 grams of water and the solutions boiling point is 101.25oC? (The KB for H2O is .51°C/m)

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24.06.2023, solved by verified expert
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The boiling point of water increases as the amount of impurities dissolved in it increases. For our purposes, we will consider the non-electrolyte to be the dissolved impurity. The change in the boiling point can be calculated using the equation:

What is the molecular mass for a non-electrolyte, №17887799, 19.11.2021 00:58

where What is the molecular mass for a non-electrolyte, №17887799, 19.11.2021 00:58 is the change in boiling point, What is the molecular mass for a non-electrolyte, №17887799, 19.11.2021 00:58 is the van ‘t Hoff factor (whose value denotes the number of particles each formula unit of the dissolved substance dissociates into in water), What is the molecular mass for a non-electrolyte, №17887799, 19.11.2021 00:58 is the boiling point elevation constant, and What is the molecular mass for a non-electrolyte, №17887799, 19.11.2021 00:58 is the molality (moles of solute/kilogram of solvent) of the solution.

Right off the bat, since we're dealing with a non-electrolyte, the dissolved substance can be assumed not to dissociate in water. So, our van ‘t Hoff factor, What is the molecular mass for a non-electrolyte, №17887799, 19.11.2021 00:58, would be 1 (by contrast, the What is the molecular mass for a non-electrolyte, №17887799, 19.11.2021 00:58 for an ionic compound like NaCl would be 2 since, in water, NaCl would dissociate into two particles: one Na⁺ ion and one Cl⁻ ion). We're also given our What is the molecular mass for a non-electrolyte, №17887799, 19.11.2021 00:58, which is 0.51 °C/m.

Assuming the normal boiling point of pure water to be 100 °C (a defined value for sig fig purposes), the change in boiling point from having dissolved 35.0 g of the non-electrolyte can be obtained by subtracting 100 °C from the final—elevated—boiling point of 101.25 °C:

What is the molecular mass for a non-electrolyte, №17887799, 19.11.2021 00:58

Now, recall what we're asked to determine: the molecular mass of the dissolved substance. There is one unknown left in the equation: the molality of the solution. Let's first solve for that:

What is the molecular mass for a non-electrolyte, №17887799, 19.11.2021 00:58

Notice that we didn't include the i since its value is 1.

Now, what would happen if we multiplied our molality by the mass of water we've been given? We would be left with the moles of solute. And what are we asked to find? The molecular mass, or the mass per mole. We can accomplish this in two steps. Remember to convert your mass of water to kilograms:

What is the molecular mass for a non-electrolyte, №17887799, 19.11.2021 00:58

And, finally, we divide the mass of our solute by the number of moles of solute:

What is the molecular mass for a non-electrolyte, №17887799, 19.11.2021 00:58

Our answer to two significant figures (which is the number of sig figs to which our What is the molecular mass for a non-electrolyte, №17887799, 19.11.2021 00:58 is given) would be 320 g/mol.

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Chemistry
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The boiling point of water increases as the amount of impurities dissolved in it increases. For our purposes, we will consider the non-electrolyte to be the dissolved impurity. The change in the boiling point can be calculated using the equation:

\Delta T_b = i \times K_b \times m

where \Delta T_b is the change in boiling point, i is the van ‘t Hoff factor (whose value denotes the number of particles each formula unit of the dissolved substance dissociates into in water), K_b is the boiling point elevation constant, and m is the molality (moles of solute/kilogram of solvent) of the solution.

Right off the bat, since we're dealing with a non-electrolyte, the dissolved substance can be assumed not to dissociate in water. So, our van ‘t Hoff factor, i, would be 1 (by contrast, the i for an ionic compound like NaCl would be 2 since, in water, NaCl would dissociate into two particles: one Na⁺ ion and one Cl⁻ ion). We're also given our K_b, which is 0.51 °C/m.

Assuming the normal boiling point of pure water to be 100 °C (a defined value for sig fig purposes), the change in boiling point from having dissolved 35.0 g of the non-electrolyte can be obtained by subtracting 100 °C from the final—elevated—boiling point of 101.25 °C:

\Delta T_b = 101.25\text{ }^o\text{C} - 100\text{ }^o\text{C} = 1.25\text{ }^o\text{C}

Now, recall what we're asked to determine: the molecular mass of the dissolved substance. There is one unknown left in the equation: the molality of the solution. Let's first solve for that:

m = \frac{\Delta T_b}{K_b} = \frac{1.25^\text{ o}\text{C}}{0.51^\text{ o}\text{C}/m} \\ m = 2.45 \text{ mol solute/kg water}.

Notice that we didn't include the i since its value is 1.

Now, what would happen if we multiplied our molality by the mass of water we've been given? We would be left with the moles of solute. And what are we asked to find? The molecular mass, or the mass per mole. We can accomplish this in two steps. Remember to convert your mass of water to kilograms:

2.45 \text{ mol solute/kg water} \times 0.045 \text{ kg water} = 0.110 \text{ mol solute.}

And, finally, we divide the mass of our solute by the number of moles of solute:

\frac{35.0 \text{ g solute}}{0.110 \text{ mol solute}} = 317.5 \text{ g/mol}

Our answer to two significant figures (which is the number of sig figs to which our K_b is given) would be 320 g/mol.

Chemistry
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P Answered by PhD

Answer:

52.6 gram

Step-by-step explanation:

It is clear by the equation 2(27+3×35.5)= 267 gm of AlCl3 reacts with 6× 80 = 480 gm of Br2 . So 29.2 gm reacts = 480× 29.2/267= 52.6 gm

Chemistry
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P Answered by Master

Calcium (Ca)(On the periodic table, ionization energy increases as you go up and to the right of the periodic table)

Calcium (Ca)(On the periodic table, ionization energy increases as you go up and to the right of the
Chemistry
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P Answered by PhD

glycoproteins

Explanation:

A positive reaction for Molisch's test is given by almost all carbohydrates (exceptions include tetroses & trioses). It can be noted that even some glycoproteins and nucleic acids give positive results for this test (since they tend to undergo hydrolysis when exposed to strong mineral acids and form monosaccharides).

Chemistry
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P Answered by PhD

Answer:

Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.

An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P×V = n×R×T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. The universal constant of ideal gases R has the same value for all gaseous substances.

Explanation:

In this case, you know:

P= 0.884 atm

V= ?

n= Answer:Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N 0.857 moles (where 28 g/mole is the molar mass of N₂, that is, the amount of mass that the substance contains in one mole.)

R=0.082Answer:Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N

T= 328 K

Replacing in the ideal gas law:

0.884 atm×V= 0.857 moles× 0.082Answer:Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N ×328 K

Solving:

Answer:Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N

V= 26.07 L

The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.

Chemistry
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P Answered by PhD
15 moles.Explanation:Hello,In this case, the undergoing chemical reaction is:Clearly, since carbon and oxygen are in a 1:1 molar ratio, 15 moles of carbon will completely react with 15 moles of oxygen, therefore 15 moles of oxygen remain as leftovers. In such a way, since carbon and carbon dioxide are also in a 1:1 molar ratio, the theoretical yield of carbon dioxide is 15 moles based on the stoichiometry:Best regards.
Chemistry
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P Answered by PhD
Answer: 25 g
Explanation: Given:
Original amount (N₀) = 100 g
Number of half-lives (n) = 11460/5730 = 2
Amount remaining (N) = ?
N = 1/2ⁿ × N₀
N = 1/2^2 × 100
N = 0.25 × 100
N = 25 g
Chemistry
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P Answered by PhD
Answer: 7.8125 g
Explanation: Given:
Original amount (N₀) = 500 g
Number of half-lives (n) = 9612/1602 = 6
Amount remaining (N) = ?
N = 1/2ⁿ × N₀
N = 1/2^6 × 500
N = 0.015625 × 500
N = 7.8125 g
Chemistry
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Answer: The product formed is potassium chloride.
Explanation:
Precipitation reaction is defined as the chemical reaction in which an insoluble salt is formed when two solutions are mixed containing soluble substances. The insoluble salt settles down at the bottom of the reaction mixture.

The chemical equation for the reaction of potassium phosphate and magnesium chloride follows (look at the picture)

2 moles of aqueous solution of potassium phosphate reacts with 3 moles of aqueous solution of magnesium chloride to produce 1 mole of solid magnesium phosphate and 6 moles of aqueous solution of potassium chloride.
Answer: The product formed is potassium chloride.
Explanation:
Precipitation reaction is defined a

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