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How many liters would 18.0 grams of water vapor occupy at 273 K and 1.00 atm?

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2 votes

Answer:

22.41 Liters

Step-by-step explanation:

Using the general law equation;

PV= nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

According to the information provided in this question,

mass of water vapor = 18g

Molar mass of water (H2O) = 1(2) + 16 = 18g.

mole = mass/molar mass

- mole (n) = 18/18 = 1mol

V = ?

T = 273K

p = 1 atm

Using PV = nRT

V = nRT/P

V = 1 × 0.0821 × 273/1

V = 22.41 Liters

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