Question

A 4.05 sample of an unknown gas at 35 C and 1.00 atm is stored in a 2.85 L flask.

Answer 1

Answer:

1.35 g/L

Explanation:

1) Use PV = nRT to determine moles of gas present in gas:

(1.10 atm) (2.55 L) = (n) (0.08206) (328 K)

n = 0.104 mol

2) Get molar mass of gas using calculated moles:

0.104 mol x M g/mol = 3.45g

M = 3.45 g / 0.104 mol = 33.17 g/mol

Density = Molar Mass x P / RT

= 33.17 g/mol x 1.10 atm / 0.0821 L mol / atm .K x 328 K

= 1.35 g /L

Answer 2

Answer:

The density of the gas is 1.42g/L and the molar mass is 32.92 g/mol

Explanation:

Mass of the unknown gas=m= 4.05g

Temperature=T= 35C

Pressure exerted upon the unknown gas= P= 1 atm

Volume of the flask=V= 2.85 L

Density=    ?  

Density= $\frac{mass of the unknown gas}{ volume of the container containing the gas}$

= $\frac{m}{V}$

=$\frac{4.05}{2.85}$

=1.421= 1.42g/L

Molar mass of the gas=M=?

Now,

As the formula is defined  

$\frac{n}{V}=\frac{P}{RT}$

n=$\frac{VP}{RT}$

(R= 0.082 L atm /K mol)

Since Temperature= 35C= (35+273) K

=308 K

n=$\frac{2.85\times 1}{0.082\times 308}$

n=$\frac{2.85}{25.26}$

n=0.1129 moles

Molar mass= M=$\frac{m}{n}$

=$\frac{4.05}{0.123}$

=32.92 g/mol

Hence the density of the gas is 1.42 g/L and the molar mass is 32.92 g/mol