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Chemistry

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What is the molality of a solution prepared by dissolving 2.00 mL of bromine (d = 3.103 g/mL) in 125 mL acelic acid, HC2H3O2 (d = 1.05 g/mL)
Choose one answer.
a. 0.301
b. 0.335
c. 0.296
d. 0.593

  • Chemistry -

    Show your work and I'll help you through it. I've dibe enough of these that you should know how to do these. The correct answer is listed.

  • Chemistry -

    3.10g/ml *1mole Br/79.0g =0.039 mol/ml
    0.039mol/ml * 2.00ml =0.0784 mol
    1.05 g/ml * 1mol/60g = 0.0175 mol/ml * 125 ml =2.18 mole


    I don't know where I am going wrong becaus ethe answers are not same????????

  • Chemistry -

    What is the molality of a solution prepared by dissolving 2.00 mL of bromine (d = 3.103 g/mL) in 125 mL acelic acid, HC2H3O2 (d = 1.05 g/mL)
    Choose one answer.
    a. 0.301
    b. 0.335
    c. 0.296
    d. 0.593
    molality = moles/kg solvent.
    2.00 mL bromine = what mass?
    2.00 mL x 3.103 g/mL = 6.206 grams. This is the solute. How many moles is that? 6.206 x (1 mole/159.81 g) = 0.0388 moles solute.

    125 mL acetic acid = what mass?
    125 mL x 1.05 g/mL = 131.25 grams. This is the solvent.
    m = moles solute/kg solvent =
    =0.0388/0.13125 = 0.29588 which rounds to 0.296 to three significant figures (which is what you are allowed with 2.00 mL and 125 mL). That's answer c.

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