How many molecules are in 6.0 moles of #NO_2#?

Question

Mateo Aguilar · Accepted Answer

There are #6.0xxN_A# individual #NO_2# molecules in such a quantity.

#N_A# represents Avogadro's number #=# #6.022140857(74)×10^23#. #N_A# is just a number; admittedly, it's a very large number. Why was it chosen? Well, it has the property that #N_A# individual #""^1H# atoms have a mass of #1# #g# PRECISELY. It is thus the link between the micro world of atoms and molecules, of which we can only conceive but about which we can theorize, with the macro world of grams and kilograms and litres and cubic metres, which we can conveniently measure. If I know the mass of an element OR a COMPOUND, I know exactly how many elemental or molecular species there are. I can measure out the mass of another reagent so that I have an equivalent amount of particles. This is how equivalent weight is calculated using Avogadro's number. In terms of #N_A#, how many atoms are there in your quantity? What type are they?

Axel Alvarado · Answer

undefined $3.61 \times 10^{24}$ molecules.

Wyatt Adkins · Answer

undefined To find the number of molecules in 6.0 moles of NO2, you can use Avogadro's constant, which is approximately $ 6.022 \times 10^{23} $ molecules per mole.

$$ \text{Number of molecules} = \text{Number of moles} \times \text{Avogadro's constant} $$

$$ \text{Number of molecules} = 6.0 \, \text{moles} \times 6.022 \times 10^{23} \, \text{molecules/mole} $$

$$ \text{Number of molecules} = 3.6132 \times 10^{24} \, \text{molecules} $$

So, there are approximately $ 3.6132 \times 10^{24} $ molecules in 6.0 moles of NO2.