# Why is it important to make sure your entire unknown has vaporized? Why is it important to put a...

## Question:

Why is it important to make sure your entire unknown has vaporized? Why is it important to put a pinhole in the aluminum "cap"? If 0.750 g of a gas occupies 265 mL at 25 degree C and 680 mm Hg, what is the molar mass of the gas?

## Molar mass:

Molar mass is an important parameter in stoichiometric calculations. It is defined as the quantity of mass of the substance per unit mole of it. It signifies the mass which is present in one mole of a particular compound.

STEP 1:

The particles tend to be immovable and begin to exert intermolecular forces on each other that are no longer negligible when a vapor is near the temperature at which the vapor would liquify. In this situation, the vapor behaves like an ideal gas. Thus, it is important to vaporize the whole of the unknown.

STEP 2:

It is important to put a pinhole, because the vapor from the unknown liquid should begin to exit the pinhole in the foil, after driving out all of the air in the flask.

STEP 3:

Given data are:

{eq}V=265 \enspace ml=0.265 \enspace L {/eq}

{eq}T=250^{\circ}c =298 \enspace k {/eq}

{eq}P=680 \enspace mm \enspace Hg = \dfrac{680}{760} = 0.895\enspace atm {/eq}

{eq}R = 0.0821 \enspace atm.L /mol.K {/eq}

STEP 4:

Using ideal gas law:

{eq}PV=nRT {/eq}

{eq}n=\dfrac{PV}{RT} {/eq}

{eq}n= \dfrac{0.895 \times 0.265}{0.0821 \times 298} =0.0097 \enspace mole {/eq}

Now, Molar mass = {eq}\dfrac{Mass}{No \enspace of \enspace moles} {/eq}

Molar mass= {eq}\dfrac{0.75}{0.0097}= 77.32 \enspace g/mol. {/eq}