Question

A sample of gas weighing 0.0286 g occupies a volume of 50 `cm^3` at 76 cmHg. and 25 degree centigrade. What is the molar mass of the gas?

Answer 1

`MM = 14.0` `"g/mol"`

Explanation:

We can find the molar mass `MM` of this gas using the equation

`MM = (dRT)/P`

where

• `d` is the density of the gas, in `"g/L"`

• `R` is the universal gas constant, equal to `0.082057("L"•"atm")/("mol"•"K")`

• `T` is the absolute temperature of the gas, in `"K"`

• `P` is the pressure of the gas, in `"atm"`

We need to convert some units to get where we need to be. Let's do the density first.

`1` `"L"` is equal to `1` `"dm"^3`, which is equal to `10^3` `"cm"^3`, so the density is

`((0.0286color(white)(l)"g")/(50cancel("cm"^3)))((10^3cancel("cm"^3))/(1cancel("dm"^3)))((1cancel("dm"^3))/(1color(white)(l)"L")) =color(red)(0.572` `color(red)("g/L"`

Degrees centigrade is the same as degrees Celsius, so let's convert this to `"K"` by adding `273`:

`T = 25^"o""C" + 273 = color(green)(298` `color(green)("K"`

Lastly, let's convert our pressure, which is in centimeters of mercury:

`76cancel("cm Hg")((10cancel("mm Hg"))/(1cancel("cm Hg")))((1color(white)(l)"atm")/(760cancel("mm Hg"))) = color(purple)(1` `color(purple)("atm"`

Now that we have all our values, let's plug them into the equation:

`MM = ((color(red)(0.572"g"/"L"))(0.082057("L"•"atm")/("mol"•"K"))(color(green)(298"K")))/(color(purple)(1"atm"))`

`= color(blue)(14.0` `color(blue)("g/mol"`