x

Subject - Chemistry:

Basics

MCQ - 441-2627

Question:

Density of a 2.05 M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is:

  1. 1.14 mol kg-1
  2. 3.28 mol kg-1
  3. 2.28 mol kg-1
  4. 3.28 mol kg-1

Correct Answer: C

Explanation:

The correct option is '3'.
Molarity of acetic acid solution = 2.05 M
Therefore the number of moles of acetic acid in 1 L solution = 2.05 mol
mass of acetic acid in 1 L solution = no. of moles x molar mass = 2.05 mol x 60 g mol-1 = 123 g
Note: molar mass of acetic acid is 60 g mol-1
Density, d = mass of solution / volume of solution
Hence:
mass of solution = density x volume of solution = 1.02 g mL-1 x 1000 mL = 1020 g
mass of solvent = mass of solution - mass of solute = 1020 - 123 = 897 g = 0.897 kg
molality, m = no. of moles / mass of solvent (in Kg) = 2.05 mol / 0.897 kg = 2.285 mol kg-1

Record Performance

645 MCQ for effective preparation of the test of Basics of Chemistry section.

Read the MCQ statement: Density of a 2.05 M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is: , keenly and apply the method you have learn through the video lessons for Basics to give the answer. Record your answer and check its correct answer and video explanation for MCQ No. 441-2627.

How to Answer

Solve the question for MCQ No. and decide which option (A through D/E) is the best choice to answer the MCQ, then click/tap the blue button to view the correct answer and it explanation.

Share This Page