Question

si un litro de oxigeno pesa 1,428g en condiciones normales,calcular el peso molecular del oxigeno

Answer 1

CN T y P

T = 273 K

P = 1 atm

V = 1 L

n = ¿?

m = 1.428 g Oxígeno

R = 0.082 (L atm / mol K)

Aplicar la ley de los gases ideales y calcular moles (n)

       V x P = n x R x T

n = ( 1 L  x  1  atm ) / ( 0.082 (L atm / mol K) x 273 K

n = 0.0446 moles

2. calcular masa molecular Mm

n = m/Mm

Mm = 1.428 g / 0.0446 mol

Mm = 32.01 g/mol

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otra forma

        1 mol  --------- 22.4 L

            x    ---------- 1 L

       x = 0.0446 moles

n = m/Mm

Mm = 1.428 g / 0.0446 mol

Mm = 32.01 g/mol