1. what is the volume of 0.900 moles of an ideal gas at 25.0° c and a pressure of 950.0 mm hg? 1.50 liters 17.6 liters 18.0 liters 34.2 liters2. what is the pressure, in mm hg, of 2.50 moles of an ideal gas if it has a volume of 50.0 liters when the temperature is 27.0° c?     84.2 mm hg    289 mm hg    617 mm hg    936 mm hg3. when 12.4 grams of kbr are dissolved in enough water to create a 170-gram solution, what is the solution's concentration, expressed as a percent by mass?     6.2% kbr    6.8% kbr    7.3% kbr    7.9% kbr

1. Using the ideal gas law, which states that PV = nRT, and remembering that absolute temperature must be used:
(950.0 mmHg)(V) = (0.900 moles)(R)(25.0 + 273.15 K)
The most convenient value of R is 0.08206 L-atm/mol-K, so we convert the pressure of 950.0 mmHg to atm by dividing by 760: 1.25 atm
(1.25 atm)(V) = (0.9)(0.08206)(298.15)
V = 17.6 L

2. Again using the ideal gas law, and converting temperature to Kelvin: 27 + 273.15 = 300.15 K
PV = nRT
P(50.0 L) = (2.50 moles)(0.08206 L-atm/mol-K)(300.15 K)
P = 1.23 atm
Then we multiply by 760 mmHg / 1 atm = 936 mmHg. This is the fourth of the choices.

3. If the complete solution has a mass of 170 grams, and 12.4 grams of it is the dissolved KBr, we simply have to divide the mass of solute (KBr) by the total mass of the solution: This gives us 12.4 grams / 170 grams = 0.0729. Multiplying by 100% gives 7.29% or approximately 7.3% KBr, which is the third of the choices.

Idk the answer sorry