Molality problems

label Chemistry
account_circle Unassigned
schedule 1 Day
account_balance_wallet $5

1. What is the mass of VOBr3 which needs to be added to 1000g water to make a 0.534m solution?

2. How much LiMnO4 needs to be added to 1000g water to make a 0.614m solution?

(If you could answer both questions that would be great! Thanks!) :)

Feb 3rd, 2015

1. mass VOBr3 = molecular weight of VOBr3 x moles of VOBr3

 1000g water (1kg) to make a 0.534m solution determines the molality of the solution

molality = moles solute / kg solvent

moles solute (VOBr3) = molality x kg solvent = 0.534 m x 1 kg = 0.534 moles

molecular weight of VOBr3 = 51+16+ (80x3) = 307 g/mol

mass = 307 g/mol x 0.534 moles = 163.9 grams of VOBr3

2. the same logic with LiMnO4

molecular weight = 7 + 55 + 4x16 = 126 g/mol

moles = 0.614 m x 1 kg = 0.614 moles

mass = 126 g/mol x 0.614 moles = 77.36 grams

Feb 3rd, 2015

Did you know? You can earn $20 for every friend you invite to Studypool!
Click here to
Refer a Friend
...
Feb 3rd, 2015
...
Feb 3rd, 2015
Jun 23rd, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer