JEE Main 2007 — Some Basic Concepts Of Chemistry Question with Solution

Option A :
6 moles of HCl(aq) produces = 3 moles of H₂(g)

$$\therefore$$ 1 mole of HCl(aq) produces = $$\frac{3}{6}$$ moles of H₂(g) = $$\frac{1}{2}$$ moles of H₂(g)

From avogadro hypothesis we know at STP 1 mole of any as occupies 22.4 L.

$$\therefore$$ $$\frac{1}{2}$$ moles of H₂(g) occupies = $$\frac{1}{2}\times 22.4$$ = 11.2 L

$$\therefore$$ Option (A) is correct.

Option B :

According to avogadro Hypothesis, volume is directly proportional to the no of moles of gases.
Only for gases, no of mole and volume relationship is possible i.e, for two gases if no of moles are same then their volume is also same. But here HCL is in aqueous form and H₂ is in gaseous form so we can't find volume of HCL(aq) using no of moles of HCL(aq).

$$\therefore$$ Option (B) is wrong.

Option C :

2 moles of Al(s) produces = 3 moles of H₂(g)

$$\therefore$$ 1 mole of Al(s) produces = $$\frac{3}{2}$$ moles of H₂(g)

From avogadro hypothesis we know at STP 1 mole of any as occupies 22.4 L.

$$\therefore$$ $$\frac{3}{2}$$ moles of H₂(g) occupies = $$\frac{3}{2}\times 22.4$$ = 33.6 L

6 moles of HCl(aq) produces = 3 moles of H₂(g)

$$\therefore$$ 1 mole of HCl(aq) produces = $$\frac{3}{6}$$ moles of H₂(g) = $$\frac{1}{2}$$ moles of H₂(g)

$$\therefore$$ $$\frac{1}{2}$$ moles of H₂(g) occupies = $$\frac{1}{2}\times 22.4$$ = 11.2 L

So from every mole of HCl(aq) that reacts only 11.2 L H₂(g) produces.

Form every mole of Al(s) produces 33.6 L H₂(g) but from every mole of HCl(aq) that reacts only 11.2 L H₂(g) produces not 33.6 L.

$$\therefore$$ Option (C) is wrong.

Option D :

2 moles of Al(s) produces = 3 moles of H₂(g)

$$\therefore$$ 1 mole of Al(s) produces = $$\frac{3}{2}$$ moles of H₂(g)

From avogadro hypothesis we know at STP 1 mole of any as occupies 22.4 L.

$$\therefore$$ $$\frac{3}{2}$$ moles of H₂(g) occupies = $$\frac{3}{2}\times 22.4$$ = 33.6 L

$$\therefore$$ Option (D) is wrong.