Question

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

Answer 1

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.

Answer 2

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????????

Answer 3

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
<b>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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