# Question #f9d12

##### 2 Answers

#### Explanation:

The trick here is to realize that the density of the gaseous element is actually five times that of **molecular oxygen**,

Your strategy here will be to determine the

You can use the **ideal gas law** equation

#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#

Here you have

*universal gas constant*, usually given as

**absolute temperature** of the gas

to show that the **molecular mass** of the molecule is **five times** that of *atomic oxygen*.

Under similar conditions for pressure and temperature, **equal number of moles** of gas occupy **the same volume**. If you take

#n = m/M_M#

to be the *number of moles* of gas present in a sample, you can say that

#PV = m/M_M * RT#

Rearrange to get

#P * M_M = m/V * RT#

which is equivalent to

#overbrace(m/V)^(color(blue)("density")) = P/(RT) * M_M#

As you know, the ratio between the *mass* of a substance and the *volume* it occupies gives you that substance's **density**. Assuming that you're dealing with *ideal gases*, the **same number of moles** of **same volume**

Now, you have

#m_(O_2)/V = P/(RT) * M_("M O"_2) -># foroxygen gas

and

#m_(X_3)/V = P/(RT) * M_("M X"_3) -># for thetriatomic molecule#"X"_3#

You know that

#overbrace(m_ (X_ 3)/color(red)(cancel(color(black)(V))))^(color(purple)("density of X"_ 3)) = 5 xx overbrace(m_ (O_2)/color(red)(cancel(color(black)(V))))^(color(darkgreen)("density of O"_2))#

#color(white)(aaaa)m_(X_3) = 5 xx m_(O_2)#

This means that the difference between these densities lies in the fact that each individual **molecule** of **five times heavier** than a molecule of

Therefore, the *molar mass* of **five times bigger** than the *molar mass* of

#M_ ("M X"_ 3) = 5 xx M_("M O"_2)#

The molar mass of oxygen gas is approximately equal to

#M_ ("M X"_3) = 5 xx "32 g mol"^(-1) = "160 g mol"^(-1)#

The molecule is **triatomic**, which of course means that it contains **atoms** of element

#M_"M X" = M_("M X"_3)/3 = "160 g mol"^(-1)/3 = color(green)(|bar(ul(color(white)(a/a)color(black)("55.3 g mol"^(-1))color(white)(a/a)|)))#

I'll leave the answer rounded to three **sig figs**.

Atomic mass(relative) =55.3 (up to one decimal place)

#### Explanation:

For the sake of convenient discussion the Symbol of the gaseous element is considered as E .

Since it is **tri-atomic** one its molecular formula will be written as

If its Relative atomic mass is **a** then its **Relative molecular mass will be 3a** .

Now by the given condition

Under the same condition of temperature and pressure

Hence we can write *Mass per unit voume* in place of *density *to represent the ratio.

*Again by Avogadro's Law It is known that under the same condition of temperature and pressure equal volume of all gases contain same number of molecules. Here we consider that under the same condition of temperature and pressure unit volume of all gases contain n molecules.*

So the relation becomes

**Dividing both numerator and denominator by**

The ratio

Inserting

we get

*Please Note that Relative Molecular mass,Relative Atomic mass are unit less quantities*

As discussed here

http://digipac.ca/chemical/molemass/moles2.htm