Question

1. The maximum number of molecules are present in:
[AIPMT-2004]
(a) 5L of N2 gas at STP
(b) 0.5g of H2 gas (c) 10g of O2 gas
(d) 15L of H2 gas at STP​

Answer 1

Answer:

I hope this is right answer

Answer 2

Formula

$\implies\mathtt{ n = \dfrac{w}{m} = \dfrac{M}{6.023 \times 10^{23} } = \dfrac{V}{22.4} }$

• n = moles

• w = mass

• m = molar mass

• M = molecules

• V = volume

Solution:-

(a)

Volume of N 2 (V)= 5 L

we know that ,

⟹ n = V / 22.4 L

⟹ n = 5 / 22.4

⟹ n = 0.22

we know that ,

⟹ n = M / 6.023 x 10²³

On rearranging ,

⟹ M = 0.22 x 6.023 x 10²³

⟹ M = 1.32 x 10²³

(b)

• Mass of H2 = 0.5 g

• Molar mass of H 2 = 2

From the formulae we can conclude that ,

⟹w / m = M / 6.023 x 10²³

On rearranging ,

⟹M = w / m x  6.023 x 10²³

⟹ M = 0.5 / 2 x  6.023 x 10²³

⟹M = 0.25 x  6.023 x 10²³

⟹ M = 1.5  x 10²³

(3)

• Mass of O2 = 10 g

• Molar mass of O2 = 32

From the formulae we can conclude that ,

⟹ w / m = M / 6.023 x 10²³

On rearranging ,

⟹ M = w / m x  6.023 x 10²³

⟹ M = 10 / 32 x  6.023 x 10²³

⟹ M = 0.3125  x  6.023 x 10²³

⟹ M = 1.88  x 10²³

(d)

From the formulae we can conclude that ,

⟹ M / 6.023 x 10²³= V / 22.4

On rearranging ,

⟹ M = 15 / 22.4  x  6.023 x 10²³

⟹ M = 0.66 x 6.023 x 10²³

⟹ M = 3.97 x 10²³

The answer is d