Question

calculate the number of molecules of oxygen gas that occupies a volume of 224 ml at 273 K and 3 atm pressure.

Answer 1

Hey!!!


You need to use PV=nRT. Use 0.224 L as V, 273 as T, 3 as P, and 0.082 as R. Your answer will give you number of moles of oxygen, which you an convert to number of molecules by multiplying moles by 6.02X10exp23.

Answer 2

Answer: $1.8005\times 10^{22}$ number of molecules of oxygen gas are present.

Explanation: Equation for Ideal gas is given by:

$PV=nRT$

Given conditions are:

T = 273K

P = 3atm

$R=0.082057\text{ L atm }mol^{-1}K^{-1}$ (known)

V = 224mL = 0.224L   (Conversion Factor: 1 L = 1000mL)

Now, to calculate the number of moles, we use the above equation. Putting values in it, we get

$(3atm)(0.224L)=(n)(0.082057\text{ L atm }mol^{-1}K^{-1})(273K)$

$n=0.0299mol$

Now, According to the mole concept

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules.

Hence, 0.0299 moles of oxygen gas will contain = $(0.0299)(6.022\times 10^{23})molecules=1.8005\times 10^{22}molecules$

Therefore, Number of molecules of oxygen gas are $1.8005\times 10^{22}molecules$