# Question 41adf

May 3, 2015

The molecular weight of the gas is 14.9 g/mol.

The approach to solving this problem is to determine the number of moles of gas present under those conditions for pressure and temperature, in that respective volume.

Once you know how many moles you have, you can use the weight of the gas to determine its molecular weight.

So, use the ideal gas law equation, $P V = n R T$, to solve for the number of moles, $n$.

$P V = n R T \implies n = \frac{P V}{R T}$

n = (0.967cancel("atm") * 3.50cancel("L"))/(0.082(cancel("atm") * cancel("L"))/("mol" * cancel("K")) * (273.15 + 35)cancel("K")) = "0.1340 moles"#

Since you know that this many moles have a weight of 2.00 g, you can determine the molecular weight to be

${M}_{M} = \text{2.00 g"/"0.1340 moles" = "14.925 g/mol}$

Rounded to three sig figs, the answer will be

${M}_{M} = \textcolor{g r e e n}{\text{14.9 g/mol}}$