# How many grams of CO(g) are there in 745 ml of the gas at 1.03 atm and 36 degrees celsius ?

##### 1 Answer

#### Explanation:

This is a pretty straightforward application of the ideal gas law equation

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

*universal gas constant*, usually given as

Notice that the question provides you with everything you need to find the *number of moles* of gas. Since you know the identity of the gas, you can then use its molar mass to find the mass of the sample.

The first thing to do here is make sure that the units **match those used in the** universal gas constant.

A quick comparison shows that you need to convert the volume from *mililiters* to liters and the temperature from *degrees Celsius* to Kelvin.

So, plug in these values and solve the ideal gas law equation for

#n = (PV)/(RT)#

#n = (1.03color(red)(cancel(color(black)("atm"))) * 745 * 10^(-3)color(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")))) * (273.15 + 36)color(red)(cancel(color(black)("K")))) = "0.03027 moles"#

Since carbon monoxide,

#0.3027color(red)(cancel(color(black)("moles"))) * "28.01 g"/(1color(red)(cancel(color(black)("mole")))) = "0.84786 g"#

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

#m = color(green)("0.85 g CO")#