Answer : The mass of oxygen present in the flask is 0.03597 grams.
Explanation :
First we have to determine the moles of
gas.
![\text{ Moles of }CO_2=\frac{\text{ Mass of }CO_2}{\text{ Molar mass of }CO_2}=(1.100g)/(44g/mole)=0.025moles](https://img.qammunity.org/2020/formulas/chemistry/college/qianw64h1j02wats006x14d9g8b1v4y4dz.png)
Now we have to calculate the moles of the oxygen gas.
Using ideal gas equation:
![PV=nRT](https://img.qammunity.org/2020/formulas/chemistry/high-school/uelah1l4d86yyc7nr57q25hwn1eullbhy3.png)
As, the moles is an additive property. So,
![PV=(n_(O_2)+n_(CO_2))RT](https://img.qammunity.org/2020/formulas/chemistry/college/6c135hl9ovo1it5rct6d47az7zf8wxyhpb.png)
where,
P = pressure of gas = 608 mmHg = 0.8 atm
(conversion used : 1 atm = 760 mmHg)
V = volume of gas = 1 L
T = temperature of gas = 373 K
= number of moles of oxygen gas = ?
= number of moles of carbon dioxide gas = 0.025 mole
R = gas constant =
![0.0821L.atmK^(-1)mol^(-1)](https://img.qammunity.org/2020/formulas/chemistry/college/gfnmmspdjut4wrci7wrq1k81x99edisnnm.png)
Now put all the given values in the ideal gas equation, we get:
![(0.8atm)* (1L)=(n_(O_2)+0.025)mole* (0.0821L.atmK^(-1)mol^(-1))* (373K)](https://img.qammunity.org/2020/formulas/chemistry/college/mffa6hioa4ismf4f1dg7vz161j2w6yja37.png)
![n_(O_2)=0.001124mole](https://img.qammunity.org/2020/formulas/chemistry/college/2vtg3a4tlkff5i12mbw3idr0up3e5gw16r.png)
Now we have to calculate the mass of oxygen gas.
![\text{Mass of }O_2=\text{Moles of }O_2* \text{Molar mass of }O_2](https://img.qammunity.org/2020/formulas/chemistry/middle-school/onv9sucspfymo1j2d1lryybahammqu1ww5.png)
![\text{Mass of }O_2=0.001124mole* 32g/mole=0.03597g](https://img.qammunity.org/2020/formulas/chemistry/college/8212mxr9bvsops0hm4mscijfftjx5g2d2x.png)
Therefore, the mass of oxygen present in the flask is 0.03597 grams.