The volume of 9.00 grams of water vapor at STP is 11.2 L.

Question

Quizplus · Accepted Answer

According to the ideal gas law, at STP, 1 mole or 18.02 grams of water vapor occupies 22.4 L of volume. To calculate the volume of 9.00 grams of water vapor, we need to use the mole concept:

9.00 g H2O vapor x (1 mol H2O vapor/ 18.02 g H2O vapor) x (22.4 L/mol) = 12.5 L H2O vapor

Therefore, the statement is false. The volume of 9.00 grams of water vapor at STP is 12.5 L, not 11.2 L.

The Volume of 9