How many grams of oxygen gas are contained in a 15 l sample at 1.02 atm and 28°c? show your work.

0.019 g.

Explanation:

where, P is the pressure of the gas in atm (P = 1.02 atm).

V is the volume of the gas in L (V = 15.0 L).

n is the no. of moles of the gas in mol (n = ??? mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 28°C + 273 = 301.0 K).

∴ n = PV/RT = (1.02 atm)(15.0 L)/(0.0821 L.atm/mol.K)(301.0 K) = 0.62 mol.

no. of moles = mass/molar mass.

∴ mass of oxygen = (no. of moles)(molar mass) = (0.62 mol)(32.0 g/mol) = 0.019 g.

it would be d. henry moseley

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