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.

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

The answer is b metal denting