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Question

When x molecules are removed from 200 mg of N​​​​​​2​​​​​O, 2.89×10-3 moles of N2O are left. x will be 

a.1020 molecules

b.1010 molecules

c.1021 molecules


Solution

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

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

No. of molecules in 1 mole = 6.022×10^(23)
No. of molecules in y 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 ie,rounded to 10^(20)
option a

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