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a student is instructed to make 1L of a 2 M solution of CaCl2 using dry salt how should he do this?
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@JoannaBlackwelder
\[Molarity = \frac{moles of solute}{liters of solvent}\]
\[2 M = \frac{moles of CaCl2}{1 L}\] How many moles of CaCl2 would you need? I think that's what you're asking
yea is it 222 g of cacl2
@iPwnBunnies
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like use up 222 g and then do the rest for the 1 L
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
ok i understand thanks a bunch!!!
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