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Chemistry 15 Online
OpenStudy (anonymous):

a student is instructed to make 1L of a 2 M solution of CaCl2 using dry salt how should he do this?

OpenStudy (anonymous):

@JoannaBlackwelder

OpenStudy (ipwnbunnies):

\[Molarity = \frac{moles of solute}{liters of solvent}\]

OpenStudy (ipwnbunnies):

\[2 M = \frac{moles of CaCl2}{1 L}\] How many moles of CaCl2 would you need? I think that's what you're asking

OpenStudy (anonymous):

yea is it 222 g of cacl2

OpenStudy (anonymous):

@iPwnBunnies

OpenStudy (anonymous):

like use up 222 g and then do the rest for the 1 L

OpenStudy (ipwnbunnies):

Correct. You need 2 moles of CaCl2 for a molarity of 2 M. The molar mass of CaCl2 is 110.9 grams per moles. (2 moles)*(110.9 grams per mole) ≈ 222 grams

OpenStudy (anonymous):

ok i understand thanks a bunch!!!

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