$34.05$ mL of phosphorus vapour weighs $0.0625$ g at $546^{o}$ C and $1.0$ bar pressure. What is the molar mass of phosphorus?

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Solution

$p = 0.1$ bar\\
$V = 34.05 m L = 34.05 \times 10^{- 3} L = 34.05 \times 10^{- 3} {d m}^{3}$ $R = 0.083 b a r {d m}^{3} k^{- 1} {m o l}^{- 1}$
$T = 546 ° C = (546 + 273) K = 819 K$
We can calculate the number of moles (n) by using the ideal gas equation : $p V = n R T$ .
Replacing n by $\frac{m}{M}$ , where, $m = m a s s = 0.0625 g$
$M = M o l e c u l a r m a s s$ , which we need to find.
$p V = \frac{m R T}{M}$
So, $M = \frac{m R T}{P V} = \frac{(0.0625 X 0.083 X 819)}{(0.1 X 34.05 \times 10^{- 3})}$
$= 1247.5 g {m o l}^{- 1}$

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Similar questions

34.05 mL of phosphorus vapour weighs 0.0625 g at $546^{\circ} C$ and 1 bar pressure. What is the molar mass of phosphorus?

34.05 m L

0.0625 g

546^{\circ} C

0.1 b a r

310^{\circ} C

775 t o r r

2.64 g d m^{- 3}

310^{\circ} C

2.64 {g dm}^{- 3}

0.850

0.5

g

39

g

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{m o l}^{- 1}

0.845

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