Question

when x molecules are removed from 200 mg of N2o, 2.89×10-3 moles of N2o are left. x will be

Answer 1

Molar mass of N2O = 44 g /mol

Mass of N2O = 200mg = 0.2g

moles of N2O present = 0.2/44 = 4.55×10^(-3) moles

number of moles of N₂O remaining = 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

As, 6.022×10^(23) molecules are present in 1 mol

No. of molecules in 1 mole = 6.022×10^(23) molecules

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

Thus  x = 9.97×10^(20) molecules

hence, 9.97×10^(20) molecules are present.

Answer 2

Answer:

here it is 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

Explanation:

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