Question

2. If 3.01 x 1020 molecules are removed from
98 mg of H SOą, then the number of moles of
H,SO, left are :-
(1) 0.1 x 10-3 (2)0.5 10-3
(3) 1.66 x 10-3
(4) 9.95 x 10-2​

Answer 1

Answer:

The number of moles of H₂SO₄ left are 0.5 * 10⁻³ mol.

Option 2) 0.5 * 10⁻³ mol

Step-by-step-explanation:

We have given that,

Mass of H₂SO₄ = 98 mg

No. of molecules removed = 3.01 * 10²⁰

We have to find the number of moles left of H₂SO₄.

Now,

Molar mass of H₂SO₄ = [ ( 2 * Atomic mass of H ) + ( Atomic mass of S ) + ( 4 * Atomic mass of O ) ]

⇒ Molar mass of H₂SO₄ = [ 2 + 32 + ( 4 * 16 ) ]

⇒ Molar mass of H₂SO₄ = ( 34 + 64 )

⇒ Molar mass of H₂SO₄ = 98 g/mol

Now, we know that,

No. of moles = ( Given mass ) / ( Molar mass )

∴ No. of moles of H₂SO₄ = ( 98 mg ) / ( 98 g/mol )

⇒ No. of moles of H₂SO₄ = 98 ÷ 98 mol mg/g

⇒ No. of moles of H₂SO₄ = 1 mol 0.001 g/g

⇒ No. of moles of H₂SO₄ = 0.001 mol

Now, we know that,

No. of molecules = No. of moles * Avogadro's number

∴ No. of molecules of H₂SO₄ = 0.001 * 6.022 * 10²³

⇒ No. of molecules of H₂SO₄ = 0.001 * 10³ * 6.022 * 10²⁰

⇒ No. of molecules of H₂SO₄ = 1 * 6.022 * 10²⁰

⇒ No. of molecules of H₂SO₄ = 6.022 * 10²⁰

Now,

No. of molecules left = No. of molecules of H₂SO₄ - No. of molecules removed

⇒ No. of molecules left = 6.022 * 10²⁰ - ( 3.01 * 10²⁰ )

⇒ No. of molecules left = 10²⁰ ( 6.022 - 3.01 )

⇒ No. of molecules left = 10²⁰ * 3.012

⇒ No. of molecules left = 3.012 * 10²⁰

Now,

No. of moles left = ( No. of molecules left ) / ( Avogadro's number )

⇒ No. of moles left = ( 3.012 * 10²⁰ ) / ( 6.022 * 10²³ )

⇒ No. of moles left = ( 3012 * 10¹⁷ ) / ( 6022 * 10²⁰ )

⇒ No. of moles left = 3012 ÷ 6022 * 10⁽¹⁷ ⁻ ²⁰⁾

⇒ No. of moles left = 3012 ÷ ( 3011 * 2 ) * 10⁻³

⇒ No. of moles left ≈ 1 ÷ 2 * 10⁻³

⇒ No. of moles left ≈ 0.5 * 10⁻³ mol

∴ The number of moles of H₂SO₄ left are 0.5 * 10⁻³ mol.