Ask Question
30 January, 22:36

A student dissolves 3.9g of aniline (C6H5NH2) in 200. mL of a solvent with a density of 1.05 g/mL. The student notices that the volume of the solvent does not change when the aniline dissolves in it. Calculate the molarity and molality of the student's solution. Be sure each of your answer entries has the correct number of significant digits. X10 How do I enter a number in scientific notation? molarity = x10 molarity=

+3
Answers (1)
  1. 31 January, 01:56
    0
    2.1 * 10⁻¹ M

    2.0 * 10⁻¹ m

    Explanation:

    Molarity

    The molar mass of aniline (solute) is 93.13 g/mol. The moles corresponding to 3.9 g are:

    3.9 g * (1 mol/93.13 g) = 0.042 mol

    The volume of the solution is 200 mL (0.200 L). The molarity of aniline is:

    M = 0.042 mol/0.200 L = 0.21 M = 2.1 * 10⁻¹ M

    Molality

    The moles of solute are 0.042 mol.

    The density of the solvent is 1.05 g/mL. The mass corresponding to 200 mL is:

    200 mL * 1.05 g/mL = 210 g = 0.210 kg

    The molality of aniline is:

    m = 0.042 mol/0.210 kg = 0.20 m = 2.0 * 10⁻¹ m
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “A student dissolves 3.9g of aniline (C6H5NH2) in 200. mL of a solvent with a density of 1.05 g/mL. The student notices that the volume of ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers