Question

34.05 mL of phosphorus vapours weigh 0.0625 g at 546 degree Celsius and 1.0 bar pressure. What is the molar mass of phosphorus ???​

Answer 1

$\huge\mathfrak{Solution}$

According to ideal gas equation,

PV = nRT

n = W/M

PV = WRT/M

M = WRT/PV

Substituting the values from the given value :

M = 0.0625(g) * 0.083(barL.K to the power -1) * 819.15(K) / 1(bar) * 34.04 * 10 to the power -3 (L)

M = 124.8 g mol to the power -1

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Answer 2

According to ideal gas equation,

PV = nRT

n = W/M

PV = WRT/M

M = WRT/PV

Substituting the values from the given value :

M = 0.0625(g) * 0.083(bar L.K to the power -1) * 819.15(K) / 1(bar) * 34.04 * 10 to the power -3 (L)

M = 124.8 g mole to the power -1