Question

Volume of 9.03x1023 atoms of Helium at S.T.P 1) 22.4L 2) 33.6L 3) 11.2L 4) 44L​

Answer 1

Answer:

The volume of 9.03×10²³ atoms of He at STP is 33.6L

Explanation:

Given,

   number of atoms of He = 9.03×10²³

 moles of He = number of atoms of He / Avogadro's number

⇒                    = 9.03×10²³ / 6.022×10²³

⇒                    = 1.499 ≈ 1.5mol

At STP, 1 mole of any substance occupies 22.4L volume

  volume of He = moles of He × 22.4L

⇒                        = 1.5×22.4 = 33.6L

Hence, the required answer is 33.6L (option 2)

Answer 2

Answer:

Volume of $9.03 * 10^{23}$ atoms of Helium at STP is 33.6 L.

Explanation:

Given number of atoms of helium = $9.03 * 10^{23}$

We know that 1 mole of any substance contains $6.022 *10^{23}$ atoms.

Number of moles = $\frac{9.03 * 10^{23}}{6.022 *10^{23}}$ = 1.5 moles

We know that the volume of 1 mole of any substance at STP = 22.4 L

∴Volume of 1.5 moles of Helium at STP = 22.4 × 1.5

                                                                 = 33.6 L.

⇒ Volume of $9.03 * 10^{23}$ atoms of Helium at STP = 33.6 L.

⇒ Option 2) is the correct answer.