Question

Calculate volume occupied by 16 g o2 at 300 k and 8.31 mpa

Answer 1

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Answer 2

The volume occupied by 16 g $O_2$ at 300 k and 8.31 mpa is $1.50\times 10^{8}L$

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = 8.31 mPa = $8.20\times 10^{-8}$ atm  $(1mPa=9.87\times 10^{-9}atm$

V= Volume of the gas = ?

T= Temperature of the gas = 300 K      

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas=$\frac{\text {given mass}}{\text {molar mass}}=\frac{16g}{32g/mol}=0.5moles$

$V=\frac{nRT}{P}=\frac{0.5\times 0.0821\times 300}{8.20\times 10^{-8}}=1.50\times 10^{8}L$

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