Question

Pre-Laboratory Questions and Exercises Due before lab begins. Answer in the space provided 1. a) Determine the number of moles of krypton gas contained in a 3.25 liter tank at a pressure 5.80 atm and a temperature of 25.5 °C. 12 5.80 An 7=25.59(+2733 299.51 Gas constant, R=0.0821 L. Atm. k mol = 0.769 mol vor of kryptan gas, v= 3.25h b) If the gas was oxygen instead of krypton, will the answer be the same? Why or why not? Different because molar mass of oxygen gas is diff than the molar mass of kyqton. 2. a) 0.190 g of a gas occupies 250.0 mL at STP. What is its molecular weight? b) What gas is it? 3. Calculate the molecular weight of a gas if 35.44 g of the gas stored in a 7.50 L tank exerts a pressure of 60.0 atm at a temperature of 35.5 °C.

Answer 1

RT t@ We Know, PV=nRT Here P = 5.80 atm V= 325b-, T-2515+273) = 298.5K R = 0.0821 L-atm mot?? 5.80 X 3.25 n - no. of moles = p = foi0821864905.) -0.769 If the pressure, temberature and volume are same, the no. of inoles of gases will be same whether it is oxygen os Krypton. Because n = por ano n does not depends on the molar mass. So in identical condition the value of 'n will be same. | W = Mass of gas = 0.190g 2@ we know, PV=nRI P = tam op PV = WRT. V = 250 mL ' = (250) L= 0·250L orm = WRI 27BK R=0.0821 L-atu motak oro M =0:190x 0.0821 * 273, M = Molar mass of the gas 1 X 0.250 = 17 g/mol 6 The gas is Noitz ( Molar. mass = 1791 met.) 3.We Knows, DV=nRT Paco atm Op PV = n RT V = 7.50L T =(35.5+273) = 308.5K 1 PV R = 0.0821 L atm motki? Op M-35.44x0:0821 X 308.50 W = 35:449 ( 60 7.50 1 M = Molar mass . = 4199 g/mot Op M = WRI