Question

find out the volume of $CO_2$ gas at STP ($0^\circ$C, 1atm) if its sample contains $5300 \times 10^{23}$ nucleons.

Answer 1

19.67 L

Answer 2

First, calculate the number of moles of $CO_2$ using the number of nucleons.

Number of nucleons in one molecule of $CO_2$ = 6 (from C) + 2 * 8 (from O) = 6 + 16 = 22

Number of moles = (Total number of nucleons) / (Number of nucleons per molecule * Avogadro's number)

Number of moles = $(5300 \times 10^{23}) / (22 \times 6.022 \times 10^{23})$

Number of moles ≈ 3.997 moles

At STP, 1 mole of any gas occupies 22.4 L.

Volume of $CO_2$ = Number of moles * 22.4 L/mole

Volume of $CO_2$ ≈ 3.997 * 22.4 L

Volume of $CO_2$ ≈ 89.53 L

However, the question states $5300 \times 10^{23}$ nucleons, which is a very large number, suggesting there was a typo and it was supposed to be particles of $CO_2$.

If the question meant $5.3 \times 10^{23}$ molecules of $CO_2$, then:

Moles of $CO_2 = (5.3 \times 10^{23}) / (6.022 \times 10^{23})$

Moles of $CO_2 \approx 0.88$ moles

Volume of $CO_2 = 0.88 \times 22.4$

Volume of $CO_2 \approx 19.712 L \approx 19.7 L$

If we use 5300 x 10^23 as number of molecules of CO2:

Moles of $CO_2 = (5300 \times 10^{23}) / (6.022 \times 10^{23}) = 880.11$

Volume of $CO_2 = 880.11 \times 22.4 = 19714.46 L$

Assuming the number of nucleons is correct, the number of moles is approximately:

$n = \frac{5300 \times 10^{23}}{22 \times 6.022 \times 10^{23}} \approx 3.997$

Volume = $3.997 \times 22.4 \approx 89.53 L$

If it was $5.3 \times 10^{25}$ nucleons

$n = \frac{5.3 \times 10^{25}}{22 \times 6.022 \times 10^{23}} \approx 39.97$

Volume = $39.97 \times 22.4 \approx 895.3 L$

If the question meant $0.53 \times 10^{23}$ molecules: $n = \frac{0.53 \times 10^{23}}{6.022 \times 10^{23}} \approx 0.088$ Volume = $0.088 \times 22.4 \approx 1.97 L$

If the question meant $53 \times 10^{23}$ molecules: $n = \frac{53 \times 10^{23}}{6.022 \times 10^{23}} \approx 8.8$ Volume = $8.8 \times 22.4 \approx 197.12 L$

Given the options, the closest answer assumes the original question contained a typo.