# Question #0ea44

##### 1 Answer

#### Answer:

#### Explanation:

Notice that the problem gives you three important parameters to work with

the volume occupied by the gasthe pressure of the gasthe temperature of the gas

This means that you can use the ideal gas law equation

#color(blue)(PV = nRT)#

to find the number of *moles of gas* found in the sample. Once you know how many moes of carbon dioxide you have, you can use its *molar mass* to determine how many grams you have.

So, the ideal gas law equation establishes a relationship between the pressure of the gas,

Here *universal gas constant* and is usually given as

#R = 0.082("atm" * "L")/("mol" * "K")#

This means that you must make sure that you're using the correct units for pressure, temperature, and volume, i.e. the units that match those used for

In your case, that means converting the volume of the gas from *mililiters* to *liters* and the temperatue of the gas from *degrees Celsius* to *Kelvin*

#450.6color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = "0.4506 L"#

and

#-50.5^@"C" + 273.15 = "226.65 K"#

Now plug in your values and solve the ideal gas law equation for

#n = (PV)/(RT)#

#n = (1.80color(red)(cancel(color(black)("atm"))) * 0.4506color(red)(cancel(color(black)("L"))))/(0.082(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 226.65color(red)(cancel(color(black)("K")))) = "0.04364 moles"#

Finally, you know that carbon dioxide's molar mass is equal to *every mole* of

In your case, the mass of the sample will be

#0.04364color(red)(cancel(color(black)("moles"))) * "44.01 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("1.92 g")#