In: Chemistry
A mixture of gases collected over water at 14oC has a total pressure of 1.198 atm and occupies 72 mL. How many grams of water escaped into the vapor phase?
The total pressure of gases over water at140 C, P = 1.198 atm
at 140 C the vapour pressure ofwater = 0.016atm
Total volume = 72.mL
= 0.072 L
Temperature = (14 +273 ) K
= 287 K
Ideal gas equation PV = nRT
so number of water moles , n = PV / RT
= ( 0.016 atm * 0.072L) / (0.08206 L.atm /mol.K*287K)
= 4.89 x 10-5 mol
molar mass of water = 18 g / mol
Therefore mass of water present in vapor state ,m = 18g /mol x 4.89 x 10-5 mol
= 8.80 x 10-4 g
8.80 x 10-4 g of water is escaps into vapor phase