Question

When x molecules are removed from 200 mg of N2O ,2.89 *10-3moles of N2O are left . X will be

1020molecules

1019 molecules

21 molecules

1021 molecules

solve it

Answer 1

may be 1021
molecule s

Answer 2

$\Huge{\mathfrak{\underline{\underline{answer}}}}$

⇒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)