Explain how you would determine the number of grams of Cu(NO3)2 that would be needed to make 1052 mL of a 2.50 M solution of Cu(NO3)2?

Explain how you would determine the mass of Na (s) needed to completely react with 25 mL of the Cu(NO3)2 solution created in #1, given the unbalanced equation below: __ Cu(NO3)2 (aq) + __ Na -> __ NaNO3 (aq) + ___ Cu (s)?

Explain how you would determine the new concentration of Cu(NO3)2 of 410 mL of distilled water was added to the solution of Cu(NO3)2 created in #1?

Explain how you would determine the freezing point of the solution created in #1, if the kբ of water is 1.853 °C/m. Assume the density of the solution is 1.00 g/mL

Explanation:

Question 1: Explain how you would determine the number of grams of Cu(NO₃)₂ that would be needed to make 1052 mL of a 2.50 M solution of Cu(NO₃)₂

1. Use the molarity (2.50M) and the volume in liters (1052mL = 1.052 liter) to calculate the number of moles of Cu(NO₃)₂, using the formula:

         ⇒ number of moles = molarity × volume in liters

2. Use the molar mass of Cu(NO₃)₂ to calculate the mass in grams using the formula:

Question 2: Explain how you would determine the mass of Na (s) needed to completely react with 25 mL of the Cu(NO₃)₂ solution created in #1, given the unbalanced equation below:

1. Balance the equation, which yields:

2. Calculate the number of moles of Cu(NO₃)₂ in 25 mL of solution, using the formula of molarity:

Subsititute with:

3. Use the theoretical mole ratio to calculate the number of moles of Na:

4. Use the atomic mass of Na to convert the number of moles into mass with the formula:

Question 3. Explain how you would determine the new concentration of Cu(NO₃)₂ if 410 mL of distilled water was added to the solution of Cu(NO₃)₂ created in #1?

1. Use the number of moles in the 1052 mL of solution, which was calculated in the first step of the question 1.

2. Add the two volumes: 410 mL + 1052 mL. This is the new volume of solution.

3. Use the formula of molarity:

Question 4. Explain how you would determine the freezing point of the solution created in #1, if the kf of water is 1.853 °C/m. Assume the density of the solution is 1.00 g/mL

1. Use the formula for the depression of the freezing point (a colligative property):

2. Kf is given: 1.853°C/m.

3. i is the van't Hoff factor.

This is used when the solute is ionic. It is the number of ions produced per mole of solute. For Cu(NO₃)₂ is its 3, because each mole of Cu(NO₃)₂ produces one mole of Cu⁺ and two moles of NO₃⁻.

4. m is the molality.

You must calculate the molality using the molarity and the density of the solution.

       m = number of moles of solute / kg of solvent.

Now that you have m, i, and Kf, just subsitute in the formula:

The freezing point of the solution created is the normal freezing point of water, 0ºC, less the depression of the freezing point of the solvent, ΔTf.

Therefore, you are done.

the answer is c i took the test