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How many grams of F are in 1.64 * 1022 molecules of XeF6 if the atomic...

How many grams of F are in 1.64 * 1022 molecules of XeF6 if the atomic mass of Xe is 131.3 g/mole, F is 19.0 g/mole, and XeF6 is 245.3 g/mole (the final answer should have the correct number of significant figures)

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1 mole of any substance contains 6.023 * 10^23 number of substances. Therefore, 1.64*10^22 molecules of XeF6 gives 1.64*10^22*1/6.023*10^23 = 0.272*10^-1= 0.0272moles of XeF6. The mass of 0.0272 moles of XeF6 = 0.0272*245.3 = 6.677grams. Again, 1 mole of XeF6 contains 6 moles of F atoms. i.e 245.3 grams of XeF6 contains 114 grams of fluorine. Therefore, 6.677 grams of XeF6 contains 6.677*114/245.3 = 3.10 grams of Fluorine. Hence, 3.10 grams of Fluorine is present in 1.64*10^22 molecules of XeF6.

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