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334 g x 1 mol/331.6 g x 6.02x10^23 molecules/mole = answer

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Q: How many molecules are in 334 g CBr4?
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How many molecules are in 325g of CBr4?

325 g of CBr4 have 5,9.10e23 molecules.


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The molecular mass of CBr4 is 12.0 + 4(79.9) = 331.6Amount of CBr4 = mass of substance / molecular mass = 393/331.6 = 1.19mol This means that a 393g pure sample contains 1.19 moles of tetrabromomethane. The Avogadro's number is 6.02 x 10^23 So, number of molecules of CBr4 = 1.19 x 6.02 x 10^23 = 7.13 x 10^23


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