Question

The volume occupied by 4g equivalent of oxygen at stp

Answer 1

hi there,

oxygen gas is O2 
It's molar mass is 32.00 g/mol 

moles O2 = mass / molar mass 
= 4 g / 32 g/mol 
= 0.125 moles 

At STP 1 mole of any ideal gas occupies a volume of 22.4 L 

volume O2 = moles O2 x 22.4 L/mol 
= 0.125 mol x 22.4 L/mol 
= 2.8 L

Answer 2

The volume of the gas is 2.79L.

Given: 4g equivalent of oxygen.

To find: We have to find its volume.

Solution:

According to the ideal gas equation-

PV=nRT

Here P is the pressure of the gas which is equal to 1atm.

V is the volume of the gas.

n is the number of moles.

The molar mass of oxygen is 32grm.

Thus the number of moles will be-

$\frac{4}{32} = 0.125$

T is the temperature that is equal to 273K.

R is the Ridberg constant which is equal to 0.082.

Thus volume will be-

$v = 0.125 \times 0.082 \times 273 \\ = 2.79$

The volume of the gas is 2.79L.