2.27 Estimate the average mass density of a sodium atom assuming its size to be about 2.5 Å. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the mass density of sodium in its crystalline phase: 970 kg/m3. Are the two densities of the same order of magnitude? If so, why?
2.27 Estimate the average mass density of a sodium atom assuming its size to be about 2.5 Å. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the mass density of sodium in its crystalline phase: 970 kg/m3. Are the two densities of the same order of magnitude? If so, why?
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1 Answer
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2.27
The diameter of the sodium atom = 2.5 Å = 2.5 * 10-10 m
Radius of the sodium atom = 1.25 * 10-10 m
Volume of the sodium atom = (4/3)(π) r3 = (4/3) * (22/7) * (1.25 * 10-10) m3 = 8.18 * 10-30
Mass of 1 mole atom of sodium = 23 g = 23 * 10-3 kg
1 mole of sodium contains 6.023 * 1023 atoms.
Hence mass of 1 sodium atom = (23 * 10-3)/ (6.023 * 1023) = 3.818 * 10-26 kg
Atomic mass density of sodium = M/V = (3.818 * 10-26)/ (8.18 * 10-30) kg/m3 = 4667.48 kg/m3
The density of sodium in its solid state is 4667.48 kg/m3 but in the crystalline phase is 970 kg/m3.
In crystalline phase there are voids in between atoms, hence the
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According to question, we can write
Total moles of gas = n = nOxygen + nOxygen =
Volume of gas = 12 × 22.4 litre = 268.8 litre = 2.688 × 105 cm3
Viscosity = Pascal . Second
=
= x = 1, x + 2y = -1, -x + z = 1
y = -1, Z = 0
Viscosity = [P1A-1T0]
1 msD = 1mm
10 vsD = 9msD
1vsD = 0.9 MsD
L.C. = 1MSD – 1VSD = 1 – 0.9 = 0.1 mm
Zero error = 4LC = 0.4 mm
Reading = MSR + VSR + correction
= 4.1 cm + 6 * .01 cm + (-0.04 cm) = (4.1 + 0.06 – 0.04) cm
= 4.12 cm = 412 * 10-2 cm
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