23) What is the volume of 2.454 x 1024 atoms of nitrogen gas at STP? [a] 1.096 x 1023 L
[b] 6.595 x 1048 L
[c] 2.241 x 1026 L
[d] 91.31L

d) V =  91.3 L

Explanation:

Given data:

Volume of nitrogen = ?

Temperature = standard = 273.15 K

Pressure = standard = 1 atm

Number of atoms of nitrogen = 2.454×10²⁴ atoms

Solution:

First of all we will calculate the number of moles of nitrogen by using Avogadro number.

1 mole = 6.022×10²³ atoms

2.454×10²⁴ atoms × 1 mol / 6.022×10²³ atoms

0.407×10¹ mol

4.07 mol

Volume of nitrogen:

PV = nRT

1 atm × V = 4.07 mol ×0.0821 atm.L /mol.K ×273.15 K

V =  91.3 atm.L /1 atm

V =  91.3 L

14,16 18 reacts with group