Answer:
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)
