Question

when x molecules are removed from 200mg of N2O, 2.89*10^-3 moles of N2O are left. then x=?

Answer 1

Molar mass of N₂O = 44
Mass of N₂O = 200mg = 0.2g
moles of N₂O present = 0.2/44 = 4.55×10^(-3) moles
Let moles of N₂O removed = x
moles of N₂O remained = 2.89×10^(-3)

Thus 4.55×10^(-3) - x = 2.89×10^(-3)
⇒ -x = 2.89×10^(-3) - 4.55×10^(-3) 
⇒ x = 4.55×10^(-3) - 2.89×10^(-3)
⇒ x = 1.66×10^(-3) moles

No. of molecules in 1 mole = 6.022×10^(23)
No. of molecules in x moles = y×6.022×10^(23) = 1.66×10^(-3)×6.022×10^(23)                                                     =9.97×10^(20) 

Thus  x = 9.97×10^(20) molecules

Answer 2

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