Question

1.5 M aqueous solution of acetic acid in water has density 0.9 g/ml. the molality of the solution will be

• 1.51 molal
• 1.85 molal
• 2.01 molal
• 1.02 molal​

Answer 1

Answer:

answer will be option B= 1.85 molal

Explanation:

Answer 2

Given info : 1.5 M aqueous solution of acetic acid in water has density 0.9 g/ml.

To find : The molality of the solution will be ...

solution : concentration of aqueous solution of acetic acid is 1.5 M , it means 1.5 moles of Acetic acid in 1000 ml of solution.

mass of acetic acid = no of moles × molar mass of acetic acid

= 1.5 × 60 = 90 g

now mass of solution = volume of solution × density of solution

= 1000 ml × 0.9 g/ml

= 900 g

mass of solvent = 900g - 90 g = 810 g

molality of solution = no of moles of solute (acetic acid)/mass of solvent in kg

= 1.5/(810/1000)

= 150/81

= 1.85 molal

Therefore the molality of the solution will be 1.85 molal.