Question

Question #1cb83

Answer 1

Here's what I got.

Explanation:

The trick here is to realize that under STP conditions, which are currently defined as a pressure of `"100 kPa"` and a temperature of `0^@"C"`, `1` mole of any ideal gas occupies `"22.723 L"`--this is known as the molar volume of agas at STP.

Now, you know that at STP, this gas has a density of `"1.7824 g L"^(-1)`. This tells you that at a pressure of `"100 kPa"` and a temperature of `0^@"C"`, `"1.7824 g"` of this gas occupies exactly `"1 L"`.

Use the molar volume of a gas at STP to find the number of moles present in the sample

`1 color(red)(cancel(color(black)("L"))) * "1 mole gas"/(22.723color(red)(cancel(color(black)("L")))) = "0.0440083 moles gas"`

To find the molar mass of the gas, you need to find the mass of exaftly `1` mole. Since you know that `0.0440083` moles have a mass of `"1.7824 g"`, you can say that

`1 color(red)(cancel(color(black)("mole"))) * "1.7824 g"/(0.0440083color(red)(cancel(color(black)("moles")))) = "40.501 g"`

Therefore, you can say that the molar mass of the gas is equal to

`color(darkgreen)(ul(color(black)("molar mass = 40.501 g mol"^(-1))))`

The answer is rounded to five sig figs, the number of sig figs you have for the density of the gas at STP.

`color(white)(a)`
SIDE NOTE More often than not, the molar volume of a gas at STP is given as `"22.414 mol L"^(-1)`, the value that corresponds to a pressure of `"1 atm"` and a temperature of `0^@"C"`.

If that's the value given to you, make sure to redo the calculations using `"22.414 L mol"^(-1)` instead of `"22.723 mol L"^(-1)`.