# Question #708e2

##### 1 Answer

320

#### Explanation:

I have fixed the question in two ways in order to make the problem soluble.

1st The gas pressure must be specified

2nd The gas density couldn't be 5.518×10 g/mL (in the original text) because that would give a molecular weight of 320·10⁴. No gas is knew to have such big molecules though being a gas.

Now, to solve the problem you have to apply perfect gas equation:

where, in the SI, *kelvin*.

Given the problem data, we can choose whatever amount of the gas. Let's say one milliliter, that is

We know the mass of one milliliter, that is

Provided we can calculate number of moles *mol* , which would correspond to the molecular weight. Let's do the magic.

The mass of each of those moles (molar mass) is:

Now we must transate to the molecular weight, by cutting dimensions and some non-significant figure (

This is unusually high for the molecule of a gaseous substance, but less than 352, the molecular weight of *uranium hexafluoride*,

As a final remark, consider that the perfect gas equation is actually exact for ideal gases only (a gas whose molecules have no hindrance, no reciprocal repulsion and attraction except when they hit each other). For a *real gas* this is only approximately true at high temperature (like 707 K) and low pressure (one atmosphere is low enough). So the result 320 is not quite exact.

I hope this would be useful.