# Question e95a0

May 26, 2015

The molar mass of the gas will be equal to 30.1 g/mol.

So, you're dealing with a gas that has a known density and for which you know the conditions for pressure and temperature. Your tool of choice will be the ideal gas law equation

$P V = n R T$

Notice that you have no information about the number of moles of gas present in the sample. However, you can write the number of moles of gas as the ratio between its mass and its molar mass

$n = \frac{m}{M} _ M$

Use this identity in the ideal gas lasw equation to get

$P V = \frac{m}{M} _ M \cdot R T$

This can be rearranged to incorporate the gas' density, which is defined as mass per unit of volume

$P V \cdot {M}_{M} = m \cdot R T \implies P \cdot {M}_{M} = {\underbrace{\frac{m}{V}}}_{\textcolor{b l u e}{\rho}} \cdot R T$

$P \cdot {M}_{M} = \rho \cdot R T \implies {M}_{M} = \frac{\rho \cdot R T}{P}$

Now just plug your values into the above equation and solve for ${M}_{M}$

M_M = (1.20"g"/cancel("L") * 0.082(cancel("atm") * cancel("L"))/("mol" * cancel("K")) * (273.15 + 25)cancel("K"))/(740/760cancel("atm")) = "30.13 g/mol"#

Rounded to three sig figs, the answer will be

${M}_{M} = \textcolor{g r e e n}{\text{30.1 g/mol}}$