Ncert Exercises & Solutions

5.10 34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °C and 0.1 bar pressure. What is the molar mass of phosphorus?

G i v e n, p = 0.1 b a r V

m L = 34.05 \times 10^{- 3} L = 34.05 \times 10^{- 3} d m^{3}

R = 0.083 b a r d m^{3} K^{- 1} m o l^{- 1}

T = 546 ° C = (546 + 273) K = 819 K

The number of moles (n) can be calculated using the ideal gas equation as:

p V = n R T

\Rightarrow n = \frac{p V}{R T}

= \frac{0.1 \times 34.05 \times 10^{- 3}}{0.083 \times 819}

= 5.01 \times 10^{- 5} m o l

Therefore, molar mass of phosphorous

= \frac{0.0625}{5.01 \times 10^{- 5}} = 1247.5 g m o l^{- 1}

Hence, the molar mass of phosphorus is 1247.5 g

m o l^{- 1}

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