Question

How many grams fo water vapor are in a 10.2 liter sample at 0.98 atmospheres and 26C?

Answer 1

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

Answer: 7.33 g