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5.10     34.05 \; mL of phosphorus vapour weighs 0.0625 g at 546 ^{o}C and 0.1 bar pressure. What is the molar mass of phosphorus?

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Given,

The volume of phosphorus vapour = 34.05mL  which weights about 0.0625g .

Temperature = 546^{\circ}C = 273+546 = 819K

Pressure =0.1 bar

Hence we apply the ideal gas equation,

PV=nRT , where the number of moles will be  n =\frac{PV}{RT}

= \frac{1bar\times (34.05\times 10^{-3} dm^3)}{0.083 bar\ dm^3\ K^{-1}Mol^{-1}\times 819K}

= 5\times 10^{-4} mol

\therefore Mass\ of\ 1\ mole = \frac{0.0625}{5\times 10^{-4}}g = 125g

Therefore the molar mass is 125g\ mol^{-1}

Posted by

Divya Prakash Singh

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