Question

What is the molarity of a solution containing 9.478 grams of RuCl3 in enough water to make 1.00 L of solution?

Answer 1

\[Molarity = {moles \over liters}\]\[{9.487gRuCl_3 \over Molar \ mass(207.429\ g \ per \ mole)} =.046\ moles \ RuCl_3\]\[{.047moles \over 1.00L} = .047M solution\]

Answer 2

The answer should be .046M I accidentally made it .047. Sorry.

Answer 3

Thank you for answering. No one was even noticing my question.

Answer 4

No problem.