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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.

Verified Answer :

meerkat18

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=nM

M(H2O)=18 grams/mole

n(H2O)=0.9810.2/0.08206(26+273)=0.43 moles

m(H2O)=0.4318=7.3 grams

Answer:

Amount of water vapor = 7.33 g

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

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

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

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