Question

How many ATOMS of nitrogen, are present in 4.42 moles of dinitrogen tetrafluoride, N2F4?

How many MOLES of fluorine are present in 9.50×1022 molecules of dinitrogen tetrafluoride?

Answer 1

1)

number of molecules of N2F4 = number of moles * Avogadro’s number

= 4.42 * 6.022*10^23 molecules

= 2.662*10^24 molecules

1 molecule of N2F4 has 2 atoms of N

So,

number of atoms of N = 2*2.662*10^24

= 4.524*10^24 atoms

Answer: 4.524*10^24 atoms

2)

moles of N2F4 = number of molecules / Avogadro’s number

= (9.50*10^22)/(6.022*10^23)

= 0.1578 moles

1 mole of N2F4 has 4 moles of F

So,

moles of F = 4*number of moles of N2F4

= 4*0.1578 moles

= 0.6312 moles

Answer: 0.631 mole