Question

When x molecules are removed from 200 mg of n2o 2.89*10-3moles of n2o left x is?

Answer 1

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Answer 2

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⇒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

= 6.022×10²³

⇒No. of molecules in x moles

= y×6.022×10²³

= 1.66×10^(-3)×6.022×10²³

=9.97×10^(20)