Question

THIS IS A THREE PART QUESTION IF YOU CAN HELP IT WOULD BE REALLY APPRECIATED SO I DONT FAIL.

1. If I add 2.65L of water to 245 grams of sodium acetate, what is the molarity of the NaC2H3O2 solution?
A. 10.8 m
B. 1.13 m
C. 1.08 m
D. 92.4 m

2. If I add 2.65 L of water to 245 grams of sodium acetate, what is the % by mass of NaC2H3O2 in this solution?
A. 8.46%
B.1.12%
C.10.8%
D.9.25%

3. If I add 2.65 L of water to 245 grams of sodium acetate, what is the mol fraction of NaC2H3O2 in this solution?
A. 0.203
B. 0.0846
C. 0.108
D. 0.0199

Answer 1

1. B = 1.13M

2. A. 8.46%

3. D = 0.0199

Step-by-step explanation:

1. Molarity of a a solution = number of moles of solute/ volume of solution in L

Number of moles of solute = mass of solute/molar mass of solute

Molar mass of NaC₂H₃O₂ = 82 g/mol, mass of NaC₂H₃O₂ = 245 g

Number of moles of NaC₂H₃O₂ = 245 g / 82 g/mol = 2.988 moles

Molarity of solution = 2.988 mols/ 2.65 L = 1.13 M

2. Percentage by mass of a substance = mass of substance /mass of solution × 100%

Mass of 2.65 L of water = density × volume

Density of water = 1 Kg/L = 1000 g/L; volume of waterb= 2.65 L

Mass of water = 1000 g/L × 2.65 L = 2650 g

Mass of solution = mass of water + mass of solute = 245 + 2650 =2895 g

Percentage by mass of NaC₂H₃O₂ = 245/2895 × 100% = 8.46%

3. Mole fraction of NaC₂H₃O₂ = moles of NaC₂H₃O₂/moles of solution

Moles of water = mass /molar mass

Mass of water = 2650 g; molar mass of water = 18 g/mol

Moles of water = 2650 g / 18 g/mol = 147.222 moles

Moles of solution = moles of solute + moles of water = 147.222 + 2.988 = 150.21

Moles of NaC₂H₃O₂ =2.988 moles

Moles fraction of NaC₂H₃O₂ =2.988/150.21 = 0.0199