At 0.0°C and 1.00 atm, 1.00 mol of a gas occupies a volume of0.0224 m^3. (a) Wha

Question

At 0.0°C and 1.00 atm, 1.00 mol of a gas occupies a volume of0.0224 m^3.
(a) What is the number density? (b) Estimate the average distancebetween the gas molecules. (c) If
the gas is nitrogen (N2), the principal component of air, what isthe total mass and mass density?
Required data for nitrogen can be found in the periodic tableinside the back cover of the text.

(A) (a) 2.69 × 1025/m^3 ; (b) 7 nm ; (c) 14.0 g, 1.36 kg/m^3
(B) (a) 1.34 × 1025/m^3 ; (b) 4 nm ; (c) 56.0 g, 3.65 kg/m^3
(C) (a) 4.01 × 1025/m^3 ; (b) 8 nm ; (c) 14.0 g, 1.25 kg/m^3
(D) (a) 2.69 × 1025/m^3 ; (b) 4 nm ; (c) 28.0 g, 1.25 kg/m^3
(E) (a) 6.22 × 1025/m^3 ; (b) 1 nm ; (c) 7.00 g, 4.16 kg/^m3

Explanation / Answer

A = Avagadro's number = 6.02 * 10E23 for a gram-mole (22.4liter = .0224 m3 Number / Volume = 6.02 * 10E23 / .0224 m3 = 2.69 *10E25 / m3 Let N be the number of molecules along the side of 1m3 Then N3 = 2.69 * 10E25    and N =2.99 * 10E8 Since N x = 1 m where x is the spacing   x = 1 m /(2.99 * 10E8) = 3.3 * 10E-9 m = 3.3 nm The mass of a hydrogen atom is 1.67 * 10E-27 kg The atomic mass of Nitrogen is 14 and since there are 2Nitrogen atoms in a Nitrogen molecule The mass of a Nitrogen molecule is 4.68 * 10E-26 kg Using the result from a above - 2.69 * 10E25 /m3 * 4.68 * 10E-26 kg = 1.26 kg /m3 If 1 m3 contains 1.26 kg then .0224m3 contains 1.26 kg / m3 * .0224m3 = .028 kg = 28 g Then N3 = 2.69 * 10E25    and N =2.99 * 10E8 Since N x = 1 m where x is the spacing   x = 1 m /(2.99 * 10E8) = 3.3 * 10E-9 m = 3.3 nm The mass of a hydrogen atom is 1.67 * 10E-27 kg The atomic mass of Nitrogen is 14 and since there are 2Nitrogen atoms in a Nitrogen molecule The mass of a Nitrogen molecule is 4.68 * 10E-26 kg Using the result from a above - 2.69 * 10E25 /m3 * 4.68 * 10E-26 kg = 1.26 kg /m3 If 1 m3 contains 1.26 kg then .0224m3 contains 1.26 kg / m3 * .0224m3 = .028 kg = 28 g

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.