Question

How many grams of water vapor (H2O) are in a 10.2 liter sample of 0.98 atmosphere and 26 C

Answer 1

Hey there!

Use the equation of Clapeyron:

T in kelvin :

26 + 273.15 => 299.15 K

R = 0.082

V = 10.2 L

P = 0.98 atm

number of moles :

P *V = n * R * T

0.98 * 10.2 = n * 0.082 * 299.15

9.996 = n * 24.5303

n = 9.996 / 24.5303

n = 0.4074 moles

Therefore:

Molar mass H2O = 18.01 g/mol

1 mole H2O ------------- 18.01 g
0.4074 moles ----------- m

m = 0.4074 * 18.01 / 1

m = 7.339 g of H2O