Question

How many grams of water vapor (H2O) are in a 10.2 liter sample at 0.98 atmospheres and 26°C? Show all work used to solve this problem.

Answer 1

Amount of water vapor = 7.33 g

Step-by-step explanation:

Given:

Volume of water vapor, V = 10.2 L

Temperature, T = 26 C

Pressure, P = 0.98 atm

To determine:

The amount of water vapor present in the above sample

Explanation:

Based on the ideal gas equation:

`PV = nRT`

where P = pressure, V = volume, n = moles, T = temperature, R = gas constant

`n = (PV)/(RT) = (0.98atm*10.2 L)/(0.0821Latm/mol-K*(26+273)K) =0.407`

`n (# moles\ of\ H2O ) = (mass\ H2O)/(molar\ mass\ H2O)\\\\mass\ H2O = n* molar\ mass = 0.407mole*18 g/mol=7.33\ g`

Answer 2

pV=nRT, where p-pressure, V-volume, n - amount of moles, n - gas constant (n=0.080206) n=pV/RT The amount of moles can be found by the equation: n=m/M where m - mass, M - molar mass So the mass could be calculated as: m=n*M M(H2O)=18 grams/mole n(H2O)=0.98*10.2/0.08206*(26+273)=0.43 moles m(H2O)=0.43*18=7.3 grams