Question

when x molecules are removed from 200 mg of N2O , 2.89 *10 to the power minus 3 mole of N2O are left .x will be 

Answer 1

Molar mass of N2O = 44 g Number of moles of N2O in 44 g = 1 Number of moles of N2O in 1g of N2O = 1 / 44 Number of moles of N2O in 200 mg or 200 × 10-3 g of N2O = 200 × 10-3 / 44 = 4.545 × 10-3 moles Number of moles of N2O left = 2.89 × 10-3 moles……(Given) Total Number of moles of N2O = 4.545 × 10-3 moles Number of moles removed of N2O = 4.545 × 10-3 – 2.89 × 10-3 = 1.655 × 10-3 moles 1 mole of N2O = 6.022 × 1023 molecules of N2O 1.655 × 10-3 moles of N2O = 1.655 × 10-3 × 6.022 × 1023 molecules of N2O = 9.96 × 1020 molecules of N2O

Answer 2

Answer:

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: