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View History Bookmarks Window Help Anderson University-CHE 1- www.saplinglearning.com/ibi.. arning Sore 2/25/2018 159PM 89/100 Grdebook PrintCaloulator Periodic Table Question 32 of 38 Sapling Learning ensity, in grams per lier, of a gas at STP if 4 44 L ofthe gas at 14.5 and 707.9 mm Hg weighs 0 872 9 g/L MacBook Air

Solution

Ans. Step 1: Given at 14.5⁰C-

            Volume, V = 4.44 L

            Temperature, T = 14.5⁰C = 287.65 K

            Pressure, P = 707.9 mm Hg = (707.9 / 760) atm = 0.93145 atm

Using Ideal gas equation:    PV = nRT      - equation 1

            Where, P = pressure in atm

            V = volume in L                   

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol^-1K^-1

            T = absolute temperature (in K) = (⁰C + 273.15) K

Putting the values in above equation-

            0.93145 atm x 4.44 L = n x (0.0821 atm L mol^-1K^-1) x 287.65 K

            Or, n = 4.135638 atm L / (23.616065 atm L mol^-1)

            Hence, n = 0.17512 mol

Therefore, moles of gas in the sample at 14.5⁰C = 0.17512 mol

# Step 2: Calculate volume of 0.17512 mol gas at STP (1 atm, 273.15K).

Putting the values in equation 1-

            1.00 atm x V = 0.17512 mol x (0.0821 atm L mol^-1K^-1) x 273.15 K

            Or, V = 3.9271736988 atm L / 1.00 atm

            Hence, V = 3.927 L

Therefore, volume of gas at STP = 3.927 L

# Step 3: So far, we have-

            Volume of gas at STP = 3.927 L

            Mass of gas (remains constant) = 0.872 g

Now,

            Density of gas at STP = Mass / Volume at STP

                                                = 0.872 g / (3.927 L)

                                                = 0.222 g/ L