Which of the following gives the molarity of a 17.0% by mass solution of sodium acetate, ch 3coona (molar mass = 82.0 g/mol) in water? the density of the solution is 1.09 g/ml. a. 2.26 × 10^-6 m b. 0.207m c. 2.07m d. 2.26m e. 2.72m

The molarity of this solution is 2.26 M (option D)

Explanation:

Step 1: Data given

Mass % = 17%

Molar mass of CH3COONa = 82.0 g/mol

Density of the solution = 1.09 g/mL

Step 2:

Assume the mass of the solution is 1.00 gram

⇒ 17.0 % CH3COONa = 0.17 grams

⇒ 83.0 % H2O = 0.83 grams

Step 3:

Density = 1.09 g/mL

Volume of the solution = total mass / density of solution

Volume of solution : 1.00 grams / 1.09 g/mL

Volume of the solution = 0.917 mL = 0.000917 L

Step 4: Calculate  number of moles of CH3COONa

Number of moles = Mass / molar mass

Number of moles CH3COONa = 0.170 grams / 82.0 g/mol

Number of moles = CH3COONa = 0.00207 moles

Step 5: Calculate molarity

Molarity = moles / volume

Molarity solution =   0.00207 / 0.000917 L

Molarity solution =  2.26 mol/L = 2.26 M

The molarity of this solution is 2.26 M (option D)

43.452

step-by-step explanation:

284 times .153 = 43.452

it is a good conductor

explanation: