# HF is a corrosive gas. At 2.0 atm and 300K, HF occupies a 4.50 L volume. How manygrams of HF are in the volume? 1. 2.74 g2. 0.573 g3. 4.82 g4. 7.3 g5. Not enough information is given.6. 15.6 g For any gas, the ideal gas law states the PV = nRT, where P= gas pressure in atmospheres, V = gas volume in liters, n= number of moles of gas present, T = gas temperature in degrees Kelvin, and R is the ideal gas constant, which is equal to 0.0821...

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For any gas, the ideal gas law states the PV = nRT, where P= gas pressure in atmospheres, V = gas volume in liters, n= number of moles of gas present, T = gas temperature in degrees Kelvin, and R is the ideal gas constant, which is equal to 0.0821 Liter * atmospheres/moles * degrees K.

Hence your problem, set up mathematically, should look like this:

2.0 atm * 4.50 L = n * 0.0821 L atm/K * 300 K

Do the multiplication on both sides of the equation to get:

9.0 = 24.63 n

Divide both sides by 24.63:

0.365 = n  so you know you have  0.365 moles of HF gas. Now, we have to convert that to grams.

Since the atomic mass of F is 19 and the atomic mass of H is 1, a single molecule of HF weighs 20 atomic mass units. This also means that one mole of HF weighs 20 grams. You have 0.365 moles of HF.

0.365 moles * 20 grams/mole = 7.3 grams.

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