Molality problems

Chemistry
Tutor: None Selected Time limit: 1 Day

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

Studypool's Notebank makes it easy to buy and sell old notes, study guides, reviews, etc.
Click to visit
The Notebank
...
Feb 3rd, 2015
...
Feb 3rd, 2015
Dec 10th, 2016
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