Exercise 1.14

Transcript

Exercise 1.14 The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å : 1 Å = 10⁻¹⁰ m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m³ of a mole of hydrogen atoms ? Given: 1Å=10^(−10) m Size of a hydrogen atom ≈0.5Å We assume that each atom is a sphere And the given size is effective radius of the atom Thus: 𝒓=𝟎.𝟓Å=𝟎.𝟓 × 𝟏𝟎^(−𝟏𝟎) 𝐦 Step 1: Volume of one atom [𝑉_(atom )=(4)/(3)𝜋𝑟^(3); 𝑉_(atom )=(4)/(3)𝜋(0.5×10^(−10))^(3); 𝑉_(atom )≈5.23599×10^(−31) m^(3)] Keep these extra digits during working. Step 2: Multiply by the number of atoms in one mole Using: [𝑁_(𝐴)≈6.022×10^(23) mol^(−1); 𝑉_(total )=𝑁_(𝐴)𝑉_(atom ); 𝑉_(total )≈(6.022×10^(23))(5.23599×10^(−31)); 𝑉_(total )≈3.1531×10^(−7) m^(3)] Step 3: Apply significant figures The radius 0.5Å has one significant figure. Therefore: 𝑉_(total )≈3×10^(−7) m^(3) (1 significant figure ) Step 1: Volume of one atom Now, 𝑉_(atom )=(4)/(3)𝜋𝑟^(3) =(4)/(3)𝜋(0.5×10^(−10))^(3) ≈5.23599 × 10^(−31) m^(3) Note: We keep extra digits while working, and we round off at the end Step 2: Multiply by the number of atoms in one mole One mole of hydrogen contains Avogadro's number (𝑁_(𝐴)) of atoms, which is approximately 6.022 ×10^(23) atoms /mol. Now, V_(total)=𝑁_(𝐴) × 𝑉 𝑉_(total )=(6.022×10^(23))×(5.236×10^(−31) m^(3)) 𝑉_(total )≈3.154×10^(−7) m^(3)[1.1.3] (A) Allen +1 Using: [𝑁_(𝐴)≈6.022×10^(23) mol^(−1); 𝑉_(total )=𝑁_(𝐴)𝑉_(atom ); 𝑉_(total )≈(6.022×10^(23))(5.23599×10^(−31)); 𝑉_(total )≈3.1531×10^(−7) m^(3)] Step 3: Apply significant figures The radius 0.5Å has one significant figure. Therefore: 𝑉_(total )≈3×10^(−7) m^(3) (1 significant figure ) =(6.022×10^(23))×(5.2359×10^(−31) m^(3)) ≈𝟑.𝟏𝟓𝟑𝟏 × 𝟏𝟎^(−𝟕) 𝒎^(𝟑) Step 3: Apply significant figures The radius 0.5Å has one significant figure. Thus, final answer would have 1 significant figure Therefore: Total volume ≈ 𝟑 ×𝟏𝟎^(−𝟕) 𝒎^(𝟑)