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Use the ideal gas law and show your work. ___0.00120_____ mol n = (0.977 atm)(0.030 L) / (0.08206 L atm/mol K) (298.15 K) = 0.00120 mol 13. Using the mass calculated in Calculation #7 and the number of moles calculated in Calculation #12, calculate the experimental molar mass of butane. Show your work. PV=nRT n=PV RT RT RT Step 3 Solve ideal gas law for number of moles n = PV / RT n = ( 3.0 n = 0.75 mol atm x 6.2 L ) / ( 0.08 L atm /mol K x 310 K) Molar Mass Equation The molar mass is the mass of a given substance (chemical element or chemical compound) divided by its amount of substance. Chemistry. Using molar volume (STP) or the ideal gas law equation, determine the molar mass, g/mole, of each of the following:1.)11.1g of a gas that has a volume of 1.50L at STP. 2.) 0.742g of a gas that has a volume 860mL at 1.28atm and 18∘C. 3.) 2.42g of a gas that has a volume 1.88L at 695mmHg and 19∘C. For example, if we have 1.00 mol of an ideal gas at 273K (0oC) and 1.0 atm (101.3 kPa). We can calculate the volume of the gas at these conditions using the ideal gas equation: This is the molar volume (V m) of the gas under the specified conditions. A gas will always expand to fill any container, so it is pointless to specify a gas volume