Molarity and the Mole Fraction (13.1.2)

Question 1

• Chemists use several different units to describe the concentration of a solute in a solution.

Answer 1

• Chemists use several different units to describe the concentration of a solute in a solution.

Question 2

• When finding the mole fraction of a solute, the number of moles of each component must be determined.

Answer 2

• When finding the mole fraction of a solute, the number of moles of each component must be determined.

Question 3

• When finding the molarity of a solute, the number of moles of solute and the total volume of solution must be determined.

Answer 3

• When finding the molarity of a solute, the number of moles of solute and the total volume of solution must be determined.

Question 4

Which statement best describes molarity, M?

Answer 4

the number of moles of solute in one liter of solution (A)

This is the correct expression for molarity.

Question 5

Suppose that you have 32.8 g of MgCl2 dissolved in 0.200 L of H2O. If the density of pure water is 1.00 g / mL, what are the mole fractions of MgCl2 and H2O, respectively? (The molar mass of MgCl2 is 95.22 g / mol and the molar mass of H2O is 18.02 g / mol.)

Answer 5

0.0302; 0.974 (B)

First step: find the number of moles of each substance:
Moles of Mg Cl2 = 32.8 g / 95.22 g / mol = 0.344 moles
Moles of H2O = 200. g / 18.02 g / mol = 11.1 moles
Note that 0.200 L of H2O = 200. mL. The mass of 200. mL water = 200. g

Second step: determine the total moles of solution.
Total moles of solution = (11.1 + 0.344) = 11.4 moles of solution.

Third step: find the mole fraction.
XMgCl2 = 0.344 / 11.4 = 0.0302
XH2O = (11.1/11.4) = 0.974

Question 6

Suppose that you added 19.8 g of MgCl2 to 0.250 L of water. If the density of pure water is 1.00 g / mL and the density of the final solution is 1.089 g / mL, what is the volume (in liters, L) of the final solution?

Answer 6

0.248 L (B)

First find the mass of water in the solution.
mass H2O
= (volume of H2O) (density of H2O) (1000 mL / L)
= (0.250 L) (1 g / mL) (1000 mL / L)
= 250 g of H2O

Now find the total mass of the solution:
total mass of solution
= mass of MgCl2 + mass of H2O
= (19.8 g MgCl2 ) + (250 g H2O)
= 269.8 = 270 g of solution.

The final volume of the solution is the mass of the solution divided by the density of the solution
= 270 g / 1.089 g / mL
= 248 mL
= 0.248 L.

Question 7

Which of the following statements about concentration is not true?

Answer 7

Concentration is valuable, informative data because it is not affected by temperature of the solution. (B)

This is not entirely correct. Some units of concentration are temperature dependent. Specifically, we learned that molarity, M, is the moles of solute per liters of solution. Because the term liters of solution refers to volume, it is temperature dependent (because density changes with temperature).

Question 8

Which of the following correctly matches the four listed expressions of concentration with their correct formulas?

Answer 8

molarity: moles of solute / liters of solution; molality: moles of solute / kilogram of solvent;
mole fraction: moles of solute / total number of moles;
percentage by mass: mass of solute / mass of solution (C)

Molarity is defined as the moles of solute per liters of solution. This is the only concentration term (that we have examined) that uses a volume unit (liters of solution). This is important because this makes this measure temperature sensitive. Molality is defined as moles of solute per kilogram of solution. Mole fraction just compares the number of moles of a substance to the total number of moles in the solution. Percentage by mass does the same comparison except with masses.

Question 9

How many grams of glucose (C6H12O6 ) would you need to prepare 1.25 L of 4% (by mass) glucose solution? Assume that the percentage by mass is the number of grams per 100 mL of solution.

Answer 9

50 g (C)

A 4% (by mass) glucose solution means that in 100 mL of solution there are 4 g of glucose. Therefore, in terms of liters, there are (4 g C6H12O6 / 100 mL) (1,000 mL / L) = 40 g / L. This particular solution has a volume of 1.25 L. Therefore, (1.25 L) (40 g of C6H12O6 / L) = 50 g C6H12O6.

Question 10

A solution is prepared by dissolving 31.7 g of MgCl2 in 0.318 L of H2O. What are the mole fractions of MgCl2 and H2O, respectively? (The molar mass of MgCl2 is 95.22 g / mol. The molar mass of H2O is 18.02 g / mol. The density of water is 1.00 g / mL.)

Answer 10

0.019, 0.981 (C)

Question 11

Suppose that you are told that you have a mass of solute dissolved in a given volume of solvent. Which of the following values is not needed in order to calculate the mole fraction of solute and the molarity of the solution?

Answer 11

the density of the solute (C)

You do not need to know the density of the solute in order to solve for the mole fraction of solute and the molarity of the solution.

Question 12

A 1.00 L aqueous solution of sucrose (C12H22O11 ) has a concentration of 1.58 M. Which of the following values is not needed to calculate the molality, m?

Answer 12

the density of the solute (A)

You do not need the density of the solute to solve for either mole fraction or molality.

Here are the steps you should take to calculate the mole fraction:
Step 1. Find the amount of solute in solution. Assume from the molarity that any given liter of solution contains 1.58 moles of sucrose solute (because moles of sucrose solute = (1 L of solution) (1.06 mol C12H22O11 / 1 L of solution) = 1.06 mol C12H22O11.
Step 2. Find the mass of sucrose by dividing the moles of sucrose by the molar mass of sucrose (the solute).
Step 3. Find the total mass of the solution by converting the volume to a mass:
mass solution = (1.00 L) (1,000 mL / L) (density of the solution (in g / mL) = g of solution.
Step 4. Subtract the mass of solute from the total mass of solution to solve for mass of solvent (and convert to kilograms if necessary).
Step 5. Calculate the molality as the moles of solute (sucrose) per mass of solvent.