The density of phosphorus vapour at and pressure is experimentally determined to be . If the molecul — States of Matter and Gaseous State Chemistry Question
Question
The density of phosphorus vapour at $327^\circ\text{C}$ and $1\text{ atm}$ pressure is experimentally determined to be $2.52\text{ g L}^{-1}$. If the molecular formula of the phosphorus vapour is assumed to be $P_x$, calculate the integer value of $x$. (Take atomic weight of $P = 31$, and $R = 0.0821\text{ L atm K}^{-1}\text{ mol}^{-1}$).
Answer: 4
💡 Solution & Explanation
Using the ideal gas density formula: $M = dRT/P$. $T = 327 + 273 = 600\text{ K}$. $M = (2.52 \times 0.0821 \times 600) / 1 = 124.13\text{ g/mol}$. The molar mass of $P_x$ is $31x$. Thus, $31x = 124.13 \implies x = 124.13 / 31 \approx 4.004$. The integer value is $4$.
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