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NISA [10]
3 years ago
10

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 o

ccupy 30.3 L under the same conditions ?
Chemistry
1 answer:
ddd [48]3 years ago
6 0

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.

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