How many grams fo water vapor are in a 10.2 liter sample at 0.98 atmospheres and 26C?
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Expert's answer
2013-08-07T10:51:14-0400
We use pV = nRT find n (moles) n = pV/RT P = 0.98 atm V = 10.2 L n = unknown right now R = 0.0821 (L*atm)/(k*mol) T = 26°C + 273 (since Kelvin = C + 273) = 299 K
Plug in numbers: (0.98 atm)(10.2 L) = n(0.0821)(299 K) Solve for n, n = 0.407 moles To get mass: (0.407 mol)*(18 g/mol) = 7.33 g of water
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