Answer:
![M=44.06(g)/(mol)](https://img.qammunity.org/2021/formulas/chemistry/college/r20sixkob41785nhe34t520ixtl8q8kr6m.png)
Step-by-step explanation:
Hello.
In this case, knowing the temperature, density and temperature of Y which behaves ideally, we can write the ideal gas equation:
![PV=nRT](https://img.qammunity.org/2021/formulas/physics/high-school/xmnfk8eqj9erqqq8x8idv0qbm03vnipq7i.png)
Whereas the moles are equal to the mass over the molar mass of Y:
![PV=(m)/(M) RT](https://img.qammunity.org/2021/formulas/chemistry/college/d94fmwqe86mm0f9pqfn03wyezjjqlbrx9p.png)
Thus, solving for the molar mass we write:
![M=(mRT)/(PV)](https://img.qammunity.org/2021/formulas/chemistry/college/88z3j3qc67kovns31fpfesv5fq4b5zdf6u.png)
Yet, since density is mass of over volume, we then write:
![M=(\rho RT)/(P)](https://img.qammunity.org/2021/formulas/chemistry/college/9od9jdefahoeluv36tdee4ef1f8nb3rb3e.png)
Considering the pressure in atm and the density in g/L:
![P=95.3kPa*(1atm)/(101.325kPa)=0.941atm\\\\\rho=1.696(g)/(dm^3) *(1dm^3)/(1L)=1.696(g)/(L)](https://img.qammunity.org/2021/formulas/chemistry/college/ihb731lqboixyee7cwyspgitmjst8k6a46.png)
Therefore, by plugging the values in, we obtain:
![M=(1.696(g)/(L) *0.082(atm*L)/(mol*K)*298K)/(0.941atm)\\\\M=44.06(g)/(mol)](https://img.qammunity.org/2021/formulas/chemistry/college/lug36bxe859mhzqsqkczckhhwmsldtacsn.png)
Thus, the gas may be propane (
) since it molar mass is 44.11 g/mol.
Best regards!