Question

When x molecules are removed from 200mg of N2O 2.89*10(-3)moles are left then x will be

Answer 1

1mg = 10^-3 g then ,

200mg = 200×10^-3g
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1 mole of NO2 contains = (14×2)+(16) = 44g/mol

then " y moles contains = 200mg = 200×10^-3g

y = 200×10^3/44

y = 4.55 × 10^-3 moles

then ,

4.55 × 10^-3 - X = 2.89 × 10^-3

- X = (2.89 × 10^-3) - (4.55 × 10^-3)

- X = -1.66 × 10^-3

X = 1.66 × 10^-3
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molar mass:- mass of the one mol of the substance is called molar mass

onother method

mass = number of moles × molar mass

200mg = number of moles × 44g/mol

number of moles = 200×10^-3/44

number of moles = 4.55 × 10^-3 moles

so 200mg of N2O contains 4.55 × 10^-3

therefore ,

4.55 × 10^-3 - X = 2.89 × 10^-3

- X = -1.66 × 10^-3

X = 1.66 × 10^-3

so your answer is 1.66 × 10^-3

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)