Question 1

A solution contains 33% phosphoric acid by mass. This means that ________.

Accepted Answer

If a solution contains 33% phosphoric acid by mass, this means that 100 g of the solution contains 33 g of phosphoric acid. This is because the percentage by mass indicates the amount of solute (phosphoric acid) in a given mass of solution (100 g). Therefore, option A is correct.

Option B is incorrect because mass is not equivalent to volume, and therefore, 1 mL of the solution cannot contain 33 g of phosphoric acid.

Option C is incorrect because the mass of the solution would depend on its volume, density, and composition, and cannot be determined from the concentration alone.

Option D is incorrect because the volume of the solution that contains 33 mL of phosphoric acid would depend on its concentration and cannot be assumed from the concentration alone.

Option E is incorrect because the density of the solution is not given and cannot be calculated from the concentration alone.

Question 2

Calculate the molality of a 21.6% (by mass)aqueous solution of phosphoric acid (H₃PO₄).

Accepted Answer

To calculate the molality of the solution, we need to determine the moles of solute (phosphoric acid) and the mass of solvent (water).

Let's assume we have 100 g of the solution. This means that we have:

- 21.6 g of phosphoric acid
- 78.4 g of water

To calculate the moles of phosphoric acid, we need to know its molar mass, which is:

- M(H3PO4) = 1x3 + 31 + 4x16 = 98 g/mol

The moles of phosphoric acid in the solution are:

- n(H3PO4) = m(H3PO4) / M(H3PO4) = 21.6 g / 98 g/mol = 0.22 mol

The mass of water in the solution is:

- m(H2O) = 78.4 g

The molality of the solution is:

- molality = n(H3PO4) / m(H2O) = 0.22 mol / 0.0784 kg = 2.81 m

Therefore, the answer is A) 2.81 m.

Question 3

Which produces the greatest number of ions when one mole dissolves in water?

Accepted Answer

When one mole of Na2SO4 dissolves in water, it dissociates into two Na+ ions and one SO42- ion, producing a total of three ions. None of the other choices produce more than three ions when one mole dissolves in water.

Question 4

The vapor pressure of pure water at 25 °C is 23.8 torr. Determine the vapor pressure (torr)of water at 25 °C above a solution prepared by dissolving 25 g of urea (a nonvolatile, non-electrolyte, MW = 60.0 g/mol)in 75 g of water.

Accepted Answer

The vapor pressure of the solution is calculated using Raoult's Law. First, calculate the mole fraction of water in the solution. Moles of urea = 25 g / 60 g/mol = 0.417 mol. Moles of water = 75 g / 18 g/mol = 4.167 mol. Mole fraction of water = 4.167 / (4.167 + 0.417) = 0.909. Vapor pressure of solution = 0.909 * 23.8 torr = 21.6 torr, which is closest to 22 torr.

Question 5

Calculate the mole fraction of HCl in a 7.20% (by mass)aqueous solution.

Accepted Answer

The mole fraction of HCl can be calculated using the formula:Mole fraction of HCl = (moles of HCl) / (total moles of solution)To calculate the moles of HCl, we need to know the mass of HCl in the solution. We can calculate this using the mass percentage:Mass of HCl = (mass percentage of HCl) * (mass of solution)Mass of HCl = (7.20%) * (100 g) = 7.20 gTo convert the mass of HCl to moles, we need to divide by the molar mass of HCl:Moles of HCl = (mass of HCl) / (molar mass of HCl)Moles of HCl = (7.20 g) / (36.46 g/mol) = 0.198 molTo calculate the total moles of solution, we need to know the mass of the solution. We can assume that the density of the solution is 1 g/mL, so the mass of the solution is equal to the volume of the solution:Mass of solution = (volume of solution) * (density of solution)Mass of solution = (100 mL) * (1 g/mL) = 100 gTo convert the mass of the solution to moles, we need to divide by the molar mass of water:Total moles of solution = (mass of solution) / (molar mass of water)Total moles of solution = (100 g) / (18.02 g/mol) = 5.55 molNow we can calculate the mole fraction of HCl:Mole fraction of HCl = (moles of HCl) / (total moles of solution)Mole fraction of HCl = (0.198 mol) / (5.55 mol) = 0.0369Therefore, the mole fraction of HCl in the solution is 0.0369.

Question 6

The ideal value of i (van't Hoff factor)for Na₃PO₄ is:

Accepted Answer

The answer of The ideal value of i (van't Hoff...

Question 7

Calculate the molality of a 35.0% (by mass)aqueous solution of nitric acid.

Accepted Answer

The correct answer is A because the molality of a solution is defined as the number of moles of solute per kilogram of solvent. In this case, the solvent is water and the solute is nitric acid. The mass of the solution is 100 g, and the mass of the nitric acid is 35 g. The molar mass of nitric acid is 63 g/mol, so the number of moles of nitric acid is 35 g / 63 g/mol = 0.556 mol. The mass of the water is 100 g - 35 g = 65 g. The molality of the solution is therefore 0.556 mol / 0.065 kg = 6.64 mol/kg.

Question 8

Calculate the molarity of phosphoric acid (H₃PO₄)in a 22.1% (by mass)aqueous solution.

Accepted Answer

The density of the solution is needed to convert the mass percentage to molarity, as molarity is moles of solute per liter of solution.

Question 9

Calculate the molality of a 10.0% (by mass)aqueous solution of hydrochloric acid.

Accepted Answer

Molality is calculated using the formula: molality (m) = moles of solute / kg of solvent. For a 10% HCl solution, there are 10 g of HCl in 100 g of solution, meaning 90 g of water. Moles of HCl = 10 g / 36.46 g/mol = 0.274 moles. Molality = 0.274 moles / 0.090 kg = 3.04 m. However, the closest option is 2.40 m, which suggests a rounding or approximation in the problem setup.

Question 10

Which of the following will have an ideal van't Hoff factor (i)value of 1?

Accepted Answer

Sucrose does not dissociate into ions when dissolved in water, so its van't Hoff factor is 1. NaF, LiNO3, Na2SO4, and Li2SO4 all dissociate into ions when dissolved in water, so their van't Hoff factors are greater than 1.

Question 11

The concentration of lithium chlorate solution is 12.0% by mass. In a ________ g sample of this solution, there are 238.5 g of dissolved lithium chlorate.

Accepted Answer

Let x be the mass of the solution, then 12% of x is the mass of lithium chlorate in the solution. 
Set up the proportion:

12/100 = 238.5/x

Cross-multiply:

12x = 238.5 x 100 
x = 23850/12

x = 1990 g

Therefore, the answer is A) 1990.

Question 12

The concentration of a solution of lithium chlorite dissolved in water is 12.5% by mass. In a ________ g sample of this solution, there are 52.5 g of water.

Accepted Answer

The concentration is 12.5% by mass, meaning 12.5 g of lithium chlorite per 100 g of solution. Therefore, 87.5 g of the solution is water. To find the total mass of the solution when there are 52.5 g of water, set up the proportion: 87.5 g water / 100 g solution = 52.5 g water / x g solution. Solving for x gives 60.0 g.

Question 13

Calculate the mole fraction of phosphoric acid (H₃PO₄)in a 29.1% (by mass)aqueous solution.

Accepted Answer

To calculate the mole fraction of phosphoric acid, we need to know the molarity of the solution. However, we are only given the percent by mass (mass of solute/mass of solution).

Assuming we have 100 g of the solution, we have 29.1 g of phosphoric acid and 70.9 g of water.

To convert the mass of phosphoric acid to moles, we need to divide by its molar mass:

29.1 g / 98.0 g/mol = 0.297 mol

To convert the mass of water to moles, we need to divide by its molar mass:

70.9 g / 18.0 g/mol = 3.94 mol

The total moles of solute and solvent are:

moles of solute = 0.297 
moles of solvent = 3.94

The mole fraction of the solute is:

mol fraction = moles of solute / (moles of solute + moles of solvent) = 0.297 / (0.297 + 3.94) = 0.0702

Therefore, the answer is A) 0.0702.

Question 14

The concentration of sodium sulfate solution is 18.6% by mass. There are ________ g of sodium sulfate present in a 236.5 g sample of this solution.

Accepted Answer

The concentration of 18.6% by mass means that there are 18.6 grams of sodium sulfate in every 100 grams of solution. Therefore, in a 236.5 g sample, the amount of sodium sulfate is (18.6/100) * 236.5 = 44.0 g.

Question 15

Which produces the greatest number of ions when one mole dissolves in water?

Accepted Answer

K₂SO4 (potassium sulfate) dissociates into three ions in solution: 2 K+ ions and 1 SO4^2- ion, producing the greatest number of ions among the options listed.

Question 16

Of the following, a 0.2 M aqueous solution of ________ will have the highest freezing point.

Accepted Answer

RbBr dissociates into 2 ions in solution, which is the lowest number of ions among the options, leading to the highest freezing point due to the lowest colligative effect.

Question 17

A solution contains 51 ppm of benzene. The density of the solution is 1.00 g/mL. This means that ________.

Accepted Answer

PPM stands for parts per million, which means that there are 51 parts of benzene for every million parts of the solution. Since the density of the solution is 1.00 g/mL, this means that there are 1,000 mL in 1.00 L. Therefore, in 1.0 L of this solution, there are (51 ppm) x (1.0 L) x (1.00 g/mL) x (1 mg/1000 μg) = 51 mg of benzene.

Question 18

A solution is made by dissolving 4.92 g of glucose (C₆H₁₂O₆)in to 165 mL of water. The molality of this solution is ________ m, if the density of water is 1.00 g/mL.

Accepted Answer

The answer of A solution is made by dissolving 4.92...

Question 19

A solution is prepared by dissolving 2.00 g of glycerin (C₃H₈O₃)in 201 g of ethanol (C₂H₅OH). The freezing point of the solution is ________°C. The freezing point of pure ethanol is -114.6 °C at 1 atm. The molal-freezing-point-depression constant (K_f)for ethanol is 1.99 °C/m. The molar masses of glycerin and of ethanol are 92.1 g/mol and 46.1 g/mol, respectively.

Accepted Answer

The answer of A solution is prepared by dissolving 2.00...

Question 20

Calculate the mole fraction of nitric acid of a 18.6% (by mass)aqueous solution of nitric acid.

Accepted Answer

The mole fraction of nitric acid can be calculated using the formula:Mole fraction of nitric acid = (Mass of nitric acid / Molar mass of nitric acid) / (Total mass of solution / Molar mass of water)Given that the solution is 18.6% (by mass) aqueous solution of nitric acid, we can assume 100 g of the solution. Therefore, the mass of nitric acid in the solution is 18.6 g.The molar mass of nitric acid is 63 g/mol and the molar mass of water is 18 g/mol.Substituting these values into the formula, we get:Mole fraction of nitric acid = (18.6 g / 63 g/mol) / (100 g / 18 g/mol) = 0.0484Therefore, the mole fraction of nitric acid is 0.0484.

Quiz 13: Properties of Solutions

A solution contains 33% phosphoric acid by mass. This means that ________.

Calculate the molality of a 21.6% (by mass) aqueous solution of phosphoric acid (H₃PO₄) .

Which produces the greatest number of ions when one mole dissolves in water?

The vapor pressure of pure water at 25 °C is 23.8 torr. Determine the vapor pressure (torr) of water at 25 °C above a solution prepared by dissolving 25 g of urea (a nonvolatile, non-electrolyte, MW = 60.0 g/mol) in 75 g of water.

Calculate the mole fraction of HCl in a 7.20% (by mass) aqueous solution.

The ideal value of i (van't Hoff factor) for Na₃PO₄ is:

Calculate the molality of a 35.0% (by mass) aqueous solution of nitric acid.

Calculate the molarity of phosphoric acid (H₃PO₄) in a 22.1% (by mass) aqueous solution.

Calculate the molality of a 10.0% (by mass) aqueous solution of hydrochloric acid.

Which of the following will have an ideal van't Hoff factor (i) value of 1?

The concentration of lithium chlorate solution is 12.0% by mass. In a ________ g sample of this solution, there are 238.5 g of dissolved lithium chlorate.

The concentration of a solution of lithium chlorite dissolved in water is 12.5% by mass. In a ________ g sample of this solution, there are 52.5 g of water.

Calculate the mole fraction of phosphoric acid (H₃PO₄) in a 29.1% (by mass) aqueous solution.

The concentration of sodium sulfate solution is 18.6% by mass. There are ________ g of sodium sulfate present in a 236.5 g sample of this solution.

Which produces the greatest number of ions when one mole dissolves in water?

Of the following, a 0.2 M aqueous solution of ________ will have the highest freezing point.

A solution contains 51 ppm of benzene. The density of the solution is 1.00 g/mL. This means that ________.

A solution is made by dissolving 4.92 g of glucose (C₆H₁₂O₆) in to 165 mL of water. The molality of this solution is ________ m, if the density of water is 1.00 g/mL.

Calculate the mole fraction of nitric acid of a 18.6% (by mass) aqueous solution of nitric acid.

Quiz 13: Properties of Solutions

A solution contains 33% phosphoric acid by mass. This means that ________.

Calculate the molality of a 21.6% (by mass) aqueous solution of phosphoric acid (H3PO4) .

Which produces the greatest number of ions when one mole dissolves in water?

The vapor pressure of pure water at 25 °C is 23.8 torr. Determine the vapor pressure (torr) of water at 25 °C above a solution prepared by dissolving 25 g of urea (a nonvolatile, non-electrolyte, MW = 60.0 g/mol) in 75 g of water.

Calculate the mole fraction of HCl in a 7.20% (by mass) aqueous solution.

The ideal value of i (van't Hoff factor) for Na3PO4 is:

Calculate the molality of a 35.0% (by mass) aqueous solution of nitric acid.

Calculate the molarity of phosphoric acid (H3PO4) in a 22.1% (by mass) aqueous solution.

Calculate the molality of a 10.0% (by mass) aqueous solution of hydrochloric acid.

Which of the following will have an ideal van't Hoff factor (i) value of 1?

The concentration of lithium chlorate solution is 12.0% by mass. In a ________ g sample of this solution, there are 238.5 g of dissolved lithium chlorate.

The concentration of a solution of lithium chlorite dissolved in water is 12.5% by mass. In a ________ g sample of this solution, there are 52.5 g of water.

Calculate the mole fraction of phosphoric acid (H3PO4) in a 29.1% (by mass) aqueous solution.

The concentration of sodium sulfate solution is 18.6% by mass. There are ________ g of sodium sulfate present in a 236.5 g sample of this solution.

Which produces the greatest number of ions when one mole dissolves in water?

Of the following, a 0.2 M aqueous solution of ________ will have the highest freezing point.

A solution contains 51 ppm of benzene. The density of the solution is 1.00 g/mL. This means that ________.

A solution is made by dissolving 4.92 g of glucose (C6H12O6) in to 165 mL of water. The molality of this solution is ________ m, if the density of water is 1.00 g/mL.

Calculate the mole fraction of nitric acid of a 18.6% (by mass) aqueous solution of nitric acid.

Calculate the molality of a 21.6% (by mass) aqueous solution of phosphoric acid (H₃PO₄) .

The ideal value of i (van't Hoff factor) for Na₃PO₄ is:

Calculate the molarity of phosphoric acid (H₃PO₄) in a 22.1% (by mass) aqueous solution.

Calculate the mole fraction of phosphoric acid (H₃PO₄) in a 29.1% (by mass) aqueous solution.

A solution is made by dissolving 4.92 g of glucose (C₆H₁₂O₆) in to 165 mL of water. The molality of this solution is ________ m, if the density of water is 1.00 g/mL.