Numerical on Molarity

1. Explain what changes and what stays the same when 1.00 L of a solution of NaCl is diluted to 1.80 L.

Ans: When a solution of NaCl (sodium chloride) is diluted from 1.00 L to 1.80 L, the amount of water in the solution increases while the amount of NaCl remains constant. This means that the concentration of NaCl in the solution decreases, because the same amount of NaCl is now dissolved in a larger volume of water.

The total amount of solute (NaCl) in the solution remains the same before and after dilution, as the solution is simply being diluted with an additional solvent (water) without any addition or removal of the solute.

However, the concentration of NaCl in the solution changes as a result of the dilution. The concentration of the solution is given by the amount of solute present per unit volume of the solution. When the same amount of solute is distributed over a larger volume, the concentration decreases.

In summary, when 1.00 L of a solution of NaCl is diluted to 1.80 L, the total amount of NaCl in the solution remains the same, but the concentration of NaCl in the solution decreases due to the increase in the volume of the solvent (water).

2. What does it mean when we say that a 200-mL sample and a 400-mL sample of a solution of salt have the same molarity? In what ways are the two samples identical? In what ways are these two samples different?

Ans: When we say that a 200-mL sample and a 400-mL sample of a solution of salt have the same molarity, it means that the two samples have the same concentration of salt molecules in moles per liter (M).

This also means that the amount of salt (in moles) in the two samples is proportional to the volume of the sample, i.e., the 400-mL sample contains twice as much salt as the 200-mL sample.

The two samples are identical in terms of their molarity, which means that they have the same concentration of salt molecules per unit volume. They are also likely to have similar physical properties such as color, density, and boiling point, assuming that they are both at the same temperature and pressure.

However, the two samples are different in terms of their total amount of salt and volume. The 400-mL sample contains twice as much salt as the 200-mL sample, and therefore has a larger total amount of salt. Similarly, the 400-mL sample has a larger volume than the 200-mL sample.

In summary, the two samples have the same concentration of salt molecules (molarity), but different amounts of salt and volumes.

3. There is about 1.0 g of calcium, as Ca2+, in 1.0 L of milk. What is the molarity of Ca2+ in milk?

Ans: To calculate the molarity of Ca2+ in milk, we need to know the number of moles of Ca2+ present in 1.0 L of milk.

First, we need to convert the mass of calcium to moles by dividing by its molar mass. The molar mass of calcium is 40.08 g/mol.

Number of moles of Ca2+ = Mass of Ca2+ / Molar mass of Ca2+

= 1.0 g / 40.08 g/mol

= 0.02493 mol

Next, we need to calculate the molarity of Ca2+ in milk by dividing the number of moles by the volume in liters.

Molarity of Ca2+ = Number of moles of Ca2+ / Volume of milk

= 0.02493 mol / 1.0 L

= 0.02493 M

Therefore, the molarity of Ca2+ in milk is 0.02493 M, which means that there are 0.02493 moles of Ca2+ per liter of milk.

4. What volume of a 1.00-M Fe(NO3)3 solution can be diluted to prepare 1.00 L of a solution with a concentration of 0.250 M?

Ans: To calculate the volume of a 1.00 M Fe(NO3)3 solution that can be diluted to prepare 1.00 L of a solution with a concentration of 0.250 M, we can use the dilution formula:

C1V1 = C2V2

where C1 and V1 are the initial concentration and volume of the Fe(NO3)3 solution, and C2 and V2 are the final concentration and volume of the diluted solution.

We can rearrange this formula to solve for V1:

V1 = (C2 x V2) / C1

Substituting the given values into the equation, we get:

V1 = (0.250 M x 1.00 L) / 1.00 M

V1 = 0.250 L

Therefore, 0.250 liters or 250 milliliters of a 1.00 M Fe(NO3)3 solution can be diluted to prepare 1.00 L of a solution with a concentration of 0.250 M.

If 0.2 L of a 0.384 M C2H5OH solution is diluted to a concentration of 0.154 M, what is the volume of the resulting solution?

Ans: To calculate the volume of the resulting solution after dilution, we can use the dilution formula:

C1V1 = C2V2

where C1 and V1 are the initial concentration and volume of the C2H5OH solution, and C2 and V2 are the final concentration and volume of the diluted solution.

We can rearrange this formula to solve for V2:

V2 = (C1 x V1) / C2

Substituting the given values into the equation, we get:

V2 = (0.384 M x 0.2 L) / 0.154 M

V2 = 0.499 L or 499 mL (rounded to three significant figures)

Therefore, the volume of the resulting solution after dilution is 0.499 L or 499 mL.

What volume of a 0.45 M C6H12O6 solution can be diluted to prepare 35 mL of a solution with a concentration of 0.025 M?

Ans: To calculate the volume of a 0.45 M C6H12O6 solution that can be diluted to prepare 35 mL of a solution with a concentration of 0.025 M, we can use the dilution formula:

C1V1 = C2V2

where C1 and V1 are the initial concentration and volume of the C6H12O6 solution, and C2 and V2 are the final concentration and volume of the diluted solution.

We can rearrange this formula to solve for V1:

V1 = (C2 x V2) / C1

Substituting the given values into the equation, we get:

V1 = (0.025 M x 0.035 L) / 0.45 M

V1 = 0.00196 L or 1.96 mL (rounded to two significant figures)

Therefore, 1.96 mL of a 0.45 M C6H12O6 solution can be diluted to prepare 35 mL of a solution with a concentration of 0.025 M.

A 2.50-L bottle of a solution of concentrated HCl was purchased for the general chemistry laboratory. The solution contained 985 g of HCl. What is the molarity of the solution?

Ans: To calculate the molarity of the solution of concentrated HCl, we need to know the number of moles of HCl present in the solution.

First, we need to convert the mass of HCl to moles by dividing by its molar mass. The molar mass of HCl is 36.46 g/mol.

Number of moles of HCl = Mass of HCl / Molar mass of HCl

= 985 g / 36.46 g/mol

= 27.00 mol

Next, we need to calculate the molarity of the solution by dividing the number of moles by the volume in liters.

Molarity of HCl = Number of moles of HCl / Volume of solution

= 27.00 mol / 2.50 L

= 10.8 M

Therefore, the molarity of the solution of concentrated HCl is 10.8 M, which means that there are 10.8 moles of HCl per liter of solution.

A solution of NaOH (molar mass 40 g mol-1) was prepared by dissolving 2 g of NaOH in 500 cm3 of water. Calculate molarity of the NaOH solution.

Ans: To calculate the molarity of the NaOH solution, we need to know the number of moles of NaOH in the solution and the volume of the solution in liters.

First, we need to convert the mass of NaOH to moles by dividing by its molar mass.

Number of moles of NaOH = Mass of NaOH / Molar mass of NaOH

= 2 g / 40 g/mol

= 0.05 mo

Next, we need to convert the volume of the solution from cm3 to L.

Volume of solution = 500 cm3 / 1000 cm3/L

= 0.5 L

Now we can calculate the molarity of the NaOH solution by dividing the number of moles by the volume in liters.

Molarity of NaOH = Number of moles of NaOH / Volume of solution

= 0.05 mol / 0.5 L

= 0.1 M

Therefore, the molarity of the NaOH solution is 0.1 M, which means that there are 0.1 moles of NaOH per liter of solution.

Calculate the quantity of anhydrous sodium carbonate required to produce 250 mL decimolar solution.

Ans: To calculate the quantity of anhydrous sodium carbonate required to produce a 250 mL decimolar solution, we need to know the molar mass of sodium carbonate and the number of moles of the compound needed to make the solution.

The molar mass of sodium carbonate (Na2CO3) is 105.99 g/mol.

To make a decimolar (0.1 M) solution, we need 0.1 moles of Na2CO3 per liter of solution

Since we want to make a 250 mL (0.25 L) solution, we need:

Number of moles of Na2CO3 = Molarity x Volume in liters

= 0.1 mol/L x 0.25 L

= 0.025 mol

Finally, we can calculate the quantity of anhydrous sodium carbonate needed using the formula:

Mass = Number of moles x Molar mass

Mass of Na2CO3 = Number of moles of Na2CO3 x Molar mass of Na2CO3

= 0.025 mol x 105.99 g/mol

= 2.65 g

Therefore, we need 2.65 grams of anhydrous sodium carbonate to produce a 250 mL decimolar solution.

Sulphuric acid is 95.8 % by mass. Calculate molarity and mole fraction of H2SO4 of density 1.91 g cm-3.

Ans: To calculate the molarity and mole fraction of H2SO4 in a solution with a density of 1.91 g/cm³ and a mass percentage of 95.8%, we need to first calculate the molar mass of H2SO4.

The molar mass of H2SO4 is 98.08 g/mol (2 x 1.01 g/mol for hydrogen, 32.06 g/mol for sulfur, and 4 x 16.00 g/mol for oxygen).

Molarity of H2SO4:

We can use the formula for molarity, which is:

Molarity = Number of moles of solute / Volume of solution in liters

To find the number of moles of H2SO4, we need to know the mass of H2SO4 in 1 liter of the solution. We can find this using the mass percentage and density of the solution:

Mass of H2SO4 in 1 liter of solution = Density x Volume x Mass percentage / 100

Mass of H2SO4 in 1 liter of solution = 1.91 g/cm³ x 1000 cm³ x 95.8 / 100

Mass of H2SO4 in 1 liter of solution = 3646.78 g

Number of moles of H2SO4 = Mass of H2SO4 / Molar mass of H2SO4

Number of moles of H2SO4 = 3646.78 g / 98.08 g/mol

Number of moles of H2SO4 = 37.19 mol

Now we can calculate the molarity of H2SO4:

Molarity of H2SO4 = Number of moles of H2SO4 / Volume of solution in liters

Molarity of H2SO4 = 37.19 mol / 1 L

Molarity of H2SO4 = 37.19 M

Mole fraction of H2SO4:

The mole fraction of H2SO4 is the number of moles of H2SO4 divided by the total number of moles in the solution:

Mole fraction of H2SO4 = Number of moles of H2SO4 / Total number of moles in solution

Total number of moles in solution can be calculated by dividing the mass of 1 liter of the solution by its molar mass:

Total number of moles in solution = Mass of solution / Molar mass of solution

The mass of 1 liter of the solution can be found using its density and volume:

Mass of solution = Density x Volume

Mass of solution = 1.91 g/cm³ x 1000 cm³

Mass of solution = 1910 g

Molar mass of solution can be calculated by adding the molar masses of its components:

Molar mass of solution = (Mass percentage of H2SO4 / 100) x Molar mass of H2SO4 + (100 - Mass percentage of H2SO4) / 100) x Molar mass of solvent

Molar mass of solution = (95.8 / 100) x 98.08 g/mol + (4.2 / 100) x (18.02 g/mol for water)

Molar mass of solution = 94.30 g/mol

Total number of moles in solution = Mass of solution / Molar mass of solution

Total number of moles in solution = 1910 g / 94.30 g/mol

Total number of moles in solution = 20.26 mol

Now we can calculate the mole fraction of H2SO4:

Mole fraction of H2SO4 = 37.19 mol / 20.