Calculation Of Molar Mass And Density Using The Ideal Gas Equation
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Calculation Of Molar Mass And Density Using The Ideal Gas Equation

Given that volume = V liters; pressure = P atmosphere; temperature = T Kelvin; Numuber of Moles = n.

Boyle's Law: V α 1/P (constant n, T)

Charle's Law: V α T (constant n, P)

Avogadro's Law: V α n (constant P, T)

Ideal Gas Equation: V α nT/P ; V = nRT/p (R is proportionality constant = 0.0821 L-atm/mol-K

(a) One Of two 1 Liter flask contains a gas of molar mass 30, the other contains a gas of molar mass 60, both at the same temperature. The pressure in flask A is X atm, and the mass of gas in the flask is 1.2 g. The pressure in flask B is 0.5X atm, and the mass of gas in that flask is 1.2 g. Which flask contains gas of molar mass 30, and which flask contains the gas of molar mass 60?
(b) The Hindenburg was a hydrogen-filled dirigible that exploded in 1937. If the Hindenburg held 2 x 105 m3 of hydrogen gas at 23oC and 1 atm. What mass of hydrogen gas was present?
(c) Which gas is the most dense at 1 atm and 298oK?
(i) CO2 (ii) N2O (iii) Cl2
(d) Why does a closed balloon filled with helium gas rise in air?
(e) Calculate the density of NO2 gas at 0.970 atm and 35oC.
(f) Calculate the molar mass of a gas if 2.50 g occupies 0.875 L at 685 torr and 35oC.
(g) The Dumas-bulb technique is used for determining the molar mass of an unknown by vaporizing the sample of liquid that boils below 100oC in a boiling water bath and then determining the mass of vapor required to fill the bulb.
Calculate the molar mass of an unknown liquid, given that its vapor mass is 1.012 g; volume of bulb is 354 cm3; pressure is 742 torr; and temperature is 99oC.

The strings: S7P1A15 (containership - mass).

The math:
Pj Problem of Interest is of type containership (mass).

(a) Let the number of moles of gas in flask A and flask B be nA and nB respectively.

From the ideal gas equation, nA = PAVA/RTA
So, nA = x(1)/(0.0821T) = x/(0.0821T).
Similarly, nB = 0.5x/(0.0821T).
So, nA/nB = [x/(0.0821T)]/[0.5x/(0.0821T)] = 1/0.5
So, nB = 0.5nA.
1.2 g of gas with molar mass 30 = 1.2/30 = 0.04 mol
1.2 g of gas with molar mass 60 = 1.2/60 = 0.02 mol
Since 0.02 = 0.5(0.04), nB = 0.02 and nA = 0.04
So, flask A contains gas with molar mass 30 and flask B contains gas with molar mass 60.

(b) 1 L = 10-3 m3
So, 2 x 105 m3 = 103(2 x 105) = 2 x 108 L
Mol of hydrogen gas present = n = PV/RT = (1 x 2 x 108)/[(0.0821)(296)] = 8.23 x 106 mols
Molar mass of H2 = 2
So, mass of 8.23 x 106 mols = 2 x 8.23 x 106 g = 16.5 x 106 = 1.65 x 104 kg.

(c) V = (nRT)/P (assuming n = 1 mole for each of the gases)
Volume of carbon dioxide = vCO2 = 1(0.0821)298/1 = 24.4658 L
Molar mass of carbon dioxide = mCO2 = 12 + 32 = 44 g
So, density of carbon dioxide = dCO2 = 44/24.4658 = 1.798 g/L.

Volume of N2O = vN2O = 1(0.0821)298/1 = 24.4658 L
Molar mass of N2O = mN2O = 14 x 2 + 16 = 44 g
So, density of N2O = dN2O = 44/24.4658 = 1.798 g/L.

Volume of Cl2 = vCl2 = 1(0.0821)298/1 = 24.4658 L
Molar mass of Cl2 = mCl2 = = 35.45 x 2 = 70.9 g
So, density of N2O = dN2O = 70.9/24.4658 = 2.898 g/L.
So, Chlorine is most dense at 1 atm and 298oC.
In general, for gas samples at the same conditions, molar mass determines density

(d) Density of helium less than density of air. In other words, the balloon weighs less than the air displaced by its volume.

(e) V = nRT/p (assuming 1 mole of nitrogen dioxide, NO2)
So, V = 1(0.0821)(308)/0.970 = 26.069 L
Molar mass of NO2 = 14 + 32 = 46 g
So, density of N)2 = 46/26.069 = 1.765 g/L.

(f) n = PV/RT
685 torr = 685/760 = 0.901 atm
So, n = 0.901(0.875)/(0.0821 x 308) = 0.788/25.287 = 0.031 mol.
Mass of 0.031 mol = 2.5 g
So, mass of 1 mole = 2.5/0.0312 = 80.1 g

(g) The following illustration is the Dumas-Bulb Diagram n = VP/RT.
V = 354 cm3 = 354 mL = 0.354 L.
P = 742 torr = 742/760 = 0.976 atm
T = 990C = 273 + 99 = 3720K
So, V = (0.354)(0.976)/(0.0821 x 372) = 0.346/30.54 =0.01133 mol.
mass of 0.01133 mole vapor = 1.012 g
So, mass of 1 mole of vapor = 1.012/0.01133 = 89.32 g.

Math

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