Question

If 5.12 g of oxygen O2 gas occupies a volume of 6.21L at a certain temperature and pressure, how many grams of oxygen gas will occupy 30.3 L under the same conditions ?

Answer 1

Answer : The mass of $O_2$ occupy 30.3 L under the same conditions will be, 24.9 grams.

Explanation :

First we have to calculate the moles of $O_2$

$\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}=\frac{5.12g}{32g/mol}=0.16mol$

Now we have to calculate the moles of $O_2$ in 30.3 L by using Avogadro's law.

Avogadro's law : It is defined as the volume of gas is directly proportional to the number of moles of gas at constant pressure and temperature.

$V\propto n$

or,

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$

where,

$V_1$ = initial volume of gas = 6.21 L

$V_2$ = final volume of gas = 30.3 L

$n_1$ = initial moles of gas = 0.16 mol

$n_2$ = final temperature of gas = ?

Now put all the given values in the above equation, we get:

$\frac{6.21L}{0.16mol}=\frac{30.3L}{n_2}$

$n_2=0.781mol$

Now we have to calculate the mass of $O_2$

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

Molar mass of $O_2$ = 32 g/mol

$\text{ Mass of }O_2=(0.781moles)\times (32g/mole)=24.9g$

Therefore, the mass of $O_2$ occupy 30.3 L under the same conditions will be, 24.9 grams.