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
3 answers
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.
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????????
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????????
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.
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.
2 answers