Question

How many grams of CO2 (MM=44 g/mol) would occupy an 8.8 L flask at 300 K and 1.1 atm?

Answer 1

Answer : The mass of $CO_2$ is 17.293 g.

Solution : Given,

Volume = 8.8 L

Temperature = 300 K

Pressure = 1.1 atm

Molar mass of $CO_2$ = 44 g/mole

Using ideal gas law :

$PV=nRT$

where,

P = pressure of gas

T = temperature of gas

V = volume of gas

R = gas constant = 0.0821 L atm/mole K

n = number of moles of gas

By rearranging the above formula,

$PV=\frac{w}{M}RT$         (where, w and M are mass and molar mass of gas)

$w=\frac{PVM}{RT}$

Now put all the given values in this formula, we get the mass of $CO_2$.

$w=\frac{(1.1atm)\times (8.8L)\times (44g/mole)}{(0.0821Latm/moleK)\times (300K)}=17.293g$

Therefore, the mass of $CO_2$ is 17.293 g.