Question #33404

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

Expert's answer

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**

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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