Q3.135CP - Chemistry: Molecular Nature Of Matter And Change

Question

Solar winds composed of free protons, electrons, andparticles bombard Earth constantly, knocking gas molecules outof the atmosphere. In this way, Earth loses about 3.0 kg of matterper second. It is estimated that the atmosphere will be gone inabout 50 billion years. Use this estimate to calculate the amount (mol) of nitrogen,which makes up 75.5 mass % of the atmosphere.

Step-by-Step Solution

Answer

Since the mass percent of N₂ in the atmosphere is 75.5%, we have 75.5g of N₂ per 100g of atmosphere. One kilogram is equal to 10³ g. The molar mass of N₂ is 28.02g/mol. The amount of nitrogen in the atmosphere is as follow:

${\text{Moles ofN}}_{\text{2}}{\text{ = 4.7304×10}}^{21}\text{g atmosphere×}\frac{{\text{75.5gN}}_{\text{2}}}{\text{100gatmosphere}}\text{×}\frac{{\text{1molN}}_{\text{2}}}{{\text{28.02gN}}_{\text{2}}}$

$Molesof{N}_2=\text{ 1.3}\times {\text{10}}^{20}mol$of N₂.

Step1. Moles of N 2

Given,

Rate of loss = 3.0 kg of matter/second

Time= 50 billion years

Mass% of N₂ = 75.5%

Since the mass percent of N₂ in the atmosphere is 75.5%, we have 75.5g of N₂ per 100g of atmosphere. One kilogram is equal to 10³ g. The molar mass of N₂ is 28.02g/mol. The amount of nitrogen in the atmosphere is as follows:

${\text{Moles ofN}}_{\text{2}}{\text{ = 4.7304×10}}^{21}\text{g atmosphere×}\frac{{\text{75.5gN}}_{\text{2}}}{\text{100gatmosphere}}\text{×}\frac{{\text{1molN}}_{\text{2}}}{{\text{28.02gN}}_{\text{2}}}$

$Molesof{N}_2=\text{ 1.3}\times {\text{10}}^{20}mol$of N₂.

Step2. Final calculation.

Since the mass percent of N₂ in the atmosphere is 75.5%, we have 75.5g of N₂ per 100g of atmosphere. One kilogram is equal to 10³ g. The molar mass of N₂ is 28.02g/mol. The amount of nitrogen in the atmosphere is as follow:

${\text{Moles ofN}}_{\text{2}}{\text{ = 4.7304×10}}^{21}\text{g atmosphere×}\frac{{\text{75.5gN}}_{\text{2}}}{\text{100gatmosphere}}\text{×}\frac{{\text{1molN}}_{\text{2}}}{{\text{28.02gN}}_{\text{2}}}$

${\text{Moles of N}}_{\text{2}}{\text{ = 1.3×10}}^{\text{20}}\text{mol}$ofN₂.