Question

34.05 mL phosphorus vapour weighs o.o625g at 546°C and 0.1 bar pressure.The molar mass of the phosphorus is
a)125.77gmol^-1
b)124.77gmol^-1
c)126.77gmol^-1
d)127.77gmol^-1​

Answer 1

p = 0.1 bar

V = 34.05 mL = 34.05 × 10–3 L = 34.05 × 10⁻³ dm³

R = 0.083 bar dm³ K⁻¹ mol⁻¹

T = 546°C = (546 + 273) K = 819 K

From the gas equation PV = w. RT / M, we get

M =  WRT/ PV

Substituting the given values in WRT/ PV

M = $\frac{0.0625}{0.1 \times 34.04}$ $\times 82.1 \times 819$

= 124.75 g/mol

So, your answer matches to b

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