A hanging titanium wire with diameter 2.0 mm (2.0 * 10-3 m) is initially 2.5 m long. When a 9 kg mass is hung from it, the wire stretches an amount 0.605 mm. A mole of titanium has a mass of 48 grams, and its density is 4.51 g/cm3. Find the approximate value of the effective spring stiffness of the interatomic bond.

The approximate value of the effective spring stiffness of the interatomic bond of titanium is 494.391 kN/m per mole.

Explanation:

Spring stiffness is given by

F = k * e

That is

Force = mass * acceleration

= 9*9.81 = 88.29 N

Extension = 0.605 mm = 6.05*10^ (-4) m

Spring stiffness = k = F/e = (88.29 N) / (6.05*10^ (-4) m) = 145.934 kN/m

Molar mass = 48 g

Density = 4.51 g/cm^3

Volume of one mole = Mass / Density = 48/4.51 = 10.64 cm^3 = 1.06*10^ (-5) m^3

Therefore a cross section of 2.0*10^ (-3) m with an area = (pi*D^2) / 4 = 3.14*10^ (-6) m^2 contains approximately 1 / (3.14*10 (-6)) moles or 0.295 moles

Therefore 0.295 moles has a stiffness value of 145.934 kN/m

Therefore, one mole will have an attractive force of 494.391 kN / (m*mole)