o Calculate the pH and pOH of 05 M HCI 6 points 19 20 An unk

o. Calculate the pH and pOH of 0.5 M HCI (6 points). 19. 20. An unknown metal crystallizes in face-centered cubic lattice. The density of the metal is determined to be 8.908 glem\' with an edge length of 352.4 pm. What is the unknown metal (14 points)? Page Score

Solution

Given,

Density of the metal = 8.908 g/cm³

Edge length (a)= 352.4 pm = 352.4 x 10 ^-10 cm (1pm = 10^-10cm)

Density is an intensive property (does not depend on quantity) , hence density of metal is equal to density of unit cell.

Mass of unit cell = Volume x density

Volume of unit cell = a³ = (352.4 x 10 ^-10 cm)³ = (3.524 x 10^-8cm)³ =  43.7631 x 10^-24 cm³

Mass of unit cell = V x d =   43.7631 x 10^-24 cm³  x  8.908 g/cm³ = 389.8417 x 10^-24 g = 3.89842 x 10^{-22 g}

Mass of a unit cell = Number of atoms in the unit cell x Mass of a single metal atom.

For fcc ,number of atoms per unit cell = 4

We know 1 mol ( Molar mass ) of the metal contains avogadro number of atoms.

Let M be the molar mass of the metal, then mass of 1 single metal atom = M / N_A ,where N_A is Avogadro number (6.022 x 10 ²³)

Mass of a unit cell = Number sof atoms in a unit cell x Mass of a single metal atom = 4 x M /N_A

ie 4 x M / (6.022 x 10 ²³) =  3.89842 x 10^-22g

M = 3.89842 x 10^{-22 x}  (6.022 x 10 ²³) /4 = 5.869 x 10¹ = 58.69 g /mol

Molar mass of the metal =  58.69 g /mol

The metal is Nickel