# What is the density of acetylene gas measured at pressure of 0.75 atm and temperature of 250°C?

##### 1 Answer

#### Explanation:

Your strategy here will be to use the **molar mass** of acetylene,

So, the ideal gas law equation looks like this

#color(blue)(PV = nRT)" "# , where

*universal gas constant*, usually given as

**always** expressed in *Kelvin*

Now, since you don't know how many **moles** of acetylene you have, you will have to work around it using the definition of **molar mass**.

A substance's molar mass tells you what the mass of **one mole** of that substance is. You can thus write molar mass,

#color(blue)(M_M = m/n)" "# , where

This means that you can express the number of moles,

#M_M = m/n implies n = m/M_M#

Plug this into the ideal gas law equation to get

#PV = m/M_M * RT" " " "color(red)(("*"))#

Now, density, *per unit of volume*.

#color(blue)(rho = m/V)" "# , where

Rearrange equation

#P * M_M = overbrace(m/V)^(color(blue)(=rho)) * RT#

This means that you have

#rho * RT = P * M_M implies rho = (P * M_M)/(RT)#

Acetylene has a molar mass of **do not** forget to convert the temperature from *degrees Celsius* to *Kelvin*

#rho = (0.75 color(red)(cancel(color(black)("atm"))) * 26.037"g"/color(red)(cancel(color(black)("mol"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 250)color(red)(cancel(color(black)("K")))) = "0.4547 g/L"#

Rounded to two sig figs, the number of sig figs you have for the temperature and pressure of the gas, the answer will be

#rho = color(green)("0.45 g/L")#